Column 1, 2, 3 contains conics, equation of tangents, and points of contact respectively.\[\]
Column I Column II Column III (I) ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ (i) $my={{m}^{2}}x+a$ (P)$\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$ (II)${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$ (ii) $y=mx+a\sqrt{{{m}^{2}}+1}$ (Q)$\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$ (III)${{y}^{2}}=4ax$ (iii )$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}$ (R)$\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$ (IV)${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$ (iv) $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}$ (S)$\left(\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)$
Which of the following options is the only INCORRECT combination? \[\]
A. (II) (iii) (P)\[\]
B. (II) (iv) (Q)\[\]
C. (I) (iii) (P)\[\]
D. (III)(i)(R)\[\]
| Column I | Column II | Column III |
| (I) ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ | (i) $my={{m}^{2}}x+a$ | (P)$\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$ |
| (II)${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$ | (ii) $y=mx+a\sqrt{{{m}^{2}}+1}$ | (Q)$\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$ |
| (III)${{y}^{2}}=4ax$ | (iii )$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}$ | (R)$\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$ |
| (IV)${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$ | (iv) $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}$ | (S)$\left(\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)$ |
Answer
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Hint: We assume the equation tangent as $y=mx+c$ where $m$ is the slope and $c$ is the $y-$ intercept. We put $y$ in the equation of each conic and use the condition on discriminant $D={{b}^{2}}-4ac=0$ for rational roots as coordinate point of contact. We find $c$ and the the point of contact using quadratic formula.\[\]
Complete step-by-step solution:
We know that the quadratic equation of the form $a{{x}^{2}}+bx+c=0$ where $a\ne 0,b,c\in R$ will have rational roots when the discriminant $D={{b}^{2}}-4ac=0$ . The roots of the equations are given by the quadratic formula
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us assume that the tangent of all the conics in the column-I of the question in slope-intercept form is
\[y=mx+c....\left( 1 \right)\]
Here $m$ is the slope and $c$ is the $y-$ intercept of the tangent. Now let us find the intercepts of the conics in column-I.\[\]
(I) The given equation of conic is
\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}+{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}+{{m}^{2}}{{x}^{2}}+{{c}^{2}}-{{a}^{2}}+2cmx=0 \\
& \Rightarrow \left( 1+{{m}^{2}} \right){{x}^{2}}+2cmx+{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we find,
\[\begin{align}
& D={{\left( 2cm \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4{{c}^{2}}{{m}^{2}}-4\left( {{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{c}^{2}}{{m}^{2}}-{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{2}} \right)=0 \\
& \Rightarrow {{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}} \\
& \Rightarrow c=\pm a\sqrt{1+{{m}^{2}}} \\
\end{align}\]
So the equation of the tangent is
\[y=mx\pm a\sqrt{{{m}^{2}}+1}\]
The coordinates of the point of contact of the conic are roots which we find by putting $c=a\sqrt{1+{{m}^{2}}},{{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}}$ from above calculation of intercept above. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-2cm \pm \sqrt{4\left( -{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}}{2\left( 1+{{m}^{2}} \right)} \\
& =\dfrac{-2\left( a\sqrt{1+{{m}^{2}}} \right)m \pm \sqrt{4\left( -{{a}^{2}}-{{a}^{2}}{{m}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}}{2\left( 1+{{m}^{2}} \right)} \\
& =\dfrac{-am}{\sqrt{1+{{m}^{2}}}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-am}{\sqrt{1+{{m}^{2}}}} \right)+a\sqrt{1+{{m}^{2}}}=\dfrac{a}{\sqrt{1+{{m}^{2}}}}\]
So the point of contact is $\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (I)(ii)(Q).
(II) The given equation of conic is
\[{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}+{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}+{{a}^{2}}{{m}^{2}}{{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
& \Rightarrow \left( 1+{{a}^{2}}{{m}^{2}} \right){{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find ,
\[\begin{align}
& D={{\left( 2{{a}^{2}}cm \right)}^{2}}-4\left( 1+{{a}^{2}}{{m}^{2}} \right)\left( {{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}-{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow 4\left( -{{a}^{2}}{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow {{a}^{2}}+{{a}^{4}}{{m}^{2}}={{a}^{2}}{{c}^{2}} \\
& \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}+1 \\
& \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}+1} \\
\end{align}\]
So the equation of the tangent is
\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}\]
The coordinates of the point of contact of the conic are roots which we find by putting \[{{c}^{2}}={{a}^{2}}{{m}^{2}}+1,{{c}^{2}}=\sqrt{{{a}^{2}}{{m}^{2}}+1}\] from above calculation of intercept. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}{{c}^{2}} \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}+1 \right) \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}+1}=\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}}\]
So the point of contact is $\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (II)(iv)(R). (I)(ii)(Q).\[\]
(III) The given equation of conic is
\[{{y}^{2}}=4ax\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{\left( mx+c \right)}^{2}}=4ax \\
& \Rightarrow {{m}^{2}}{{x}^{2}}+{{c}^{2}}+2cmx-4ax=0 \\
& \Rightarrow {{m}^{2}}{{x}^{2}}+\left( 2cm-4a \right)x+{{c}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we have,
\[\begin{align}
& D={{\left( 2cm-4a \right)}^{2}}-4{{m}^{2}}{{c}^{2}}=0 \\
& \Rightarrow 4{{c}^{2}}{{m}^{2}}+16{{a}^{2}}-16cma-4c{{m}^{2}}=0 \\
& \Rightarrow 16{{a}^{2}}-16cma=0 \\
& \Rightarrow c=\dfrac{a}{m} \\
\end{align}\]
So the equation of the tangent is
\[\begin{align}
& y=mx+\dfrac{a}{m} \\
& \Rightarrow my={{m}^{2}}x+a \\
\end{align}\]
The coordinates of the point of contact of the conic are roots which we find by putting $c=\dfrac{a}{m},{{c}^{2}}=\dfrac{{{a}^{2}}}{{{m}^{2}}}$ from calculation of intercept above. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-\left( 2cm-4a \right)\pm\sqrt{16{{a}^{2}}-16cma}}{2{{m}^{2}}} \\
& =\dfrac{-2\left( \dfrac{a}{m} \right)m+4a \pm \sqrt{16{{a}^{2}}-16\left( \dfrac{a}{m} \right)ma}}{2{{m}^{2}}} \\
& =\dfrac{a}{{{m}^{2}}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{a}{{{m}^{2}}} \right)+\dfrac{a}{m}=\dfrac{2a}{m}\]
So the point of contact is $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$. We match the columns with the obtained result and find the matching as (III)(i)(P). (II)(iv)(R). (I)(ii)(Q). .\[\]
(IV) The given equation of conic is
\[{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}-{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}-{{a}^{2}}{{m}^{2}}{{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
&\Rightarrow \left( 1-{{a}^{2}}{{m}^{2}} \right){{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find,
\[\begin{align}
& D={{\left( -2{{a}^{2}}cm \right)}^{2}}-4\left( 1-{{a}^{2}}{{m}^{2}} \right)\left( -{{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow {{a}^{2}}-{{a}^{4}}{{m}^{2}}=-{{a}^{2}}{{c}^{2}} \\
& \Rightarrow 1-{{a}^{2}}{{m}^{2}}=-{{c}^{2}} \\
& \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}-1 \\
& \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}-1} \\
\end{align}\]
So the equation of the tangent is
\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}\]
The coordinates of the point of contact of the conic are roots which we find by putting \[{{c}^{2}}={{a}^{2}}{{m}^{2}}-1,c=\sqrt{{{a}^{2}}{{m}^{2}}-1}\] from above calculation of intercept. We have the $x-$coordinate of point of contact as,
contact as,
\[\begin{align}
& x=\dfrac{2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}{{c}^{2}} \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\
& =\dfrac{2{{a}^{2}}m\left( \sqrt{{{a}^{2}}{{m}^{2}}-1} \right)\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}-1 \right) \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}-1}}{2\left( {{a}^{2}}{{m}^{2}}-1 \right)} \\
& =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}-1}=\dfrac{-1}{\sqrt{1+{{m}^{2}}}}\]
So the point of contact is \[\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)\] . We match the columns with the obtained result and find the matching as (IV)(iii)(S).
So the correct matches are (IV)(iii)(S), (III)(i)(P), (II)(iv)(R), (I)(ii)(Q). We are asked to choose the only INCORRECT combination in the options which is all of the options A, B,C,D.
Note: We have taken negative and positive square root for $c$ keeping the options in mind .We note that the equation ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is an equation of circle in centre-radius from, ${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of ellipse in axis form, ${{y}^{2}}=4ax$ is an equation of parabola in vertex from and ${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of hyperbola in axis from. We can directly find tangents and point of contact if we remember the formulae.
Complete step-by-step solution:
We know that the quadratic equation of the form $a{{x}^{2}}+bx+c=0$ where $a\ne 0,b,c\in R$ will have rational roots when the discriminant $D={{b}^{2}}-4ac=0$ . The roots of the equations are given by the quadratic formula
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us assume that the tangent of all the conics in the column-I of the question in slope-intercept form is
\[y=mx+c....\left( 1 \right)\]
Here $m$ is the slope and $c$ is the $y-$ intercept of the tangent. Now let us find the intercepts of the conics in column-I.\[\]
(I) The given equation of conic is
\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}+{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}+{{m}^{2}}{{x}^{2}}+{{c}^{2}}-{{a}^{2}}+2cmx=0 \\
& \Rightarrow \left( 1+{{m}^{2}} \right){{x}^{2}}+2cmx+{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we find,
\[\begin{align}
& D={{\left( 2cm \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4{{c}^{2}}{{m}^{2}}-4\left( {{c}^{2}}-{{a}^{2}}+{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{c}^{2}}{{m}^{2}}-{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{2}} \right)=0 \\
& \Rightarrow {{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}} \\
& \Rightarrow c=\pm a\sqrt{1+{{m}^{2}}} \\
\end{align}\]
So the equation of the tangent is
\[y=mx\pm a\sqrt{{{m}^{2}}+1}\]
The coordinates of the point of contact of the conic are roots which we find by putting $c=a\sqrt{1+{{m}^{2}}},{{c}^{2}}={{a}^{2}}+{{a}^{2}}{{m}^{2}}$ from above calculation of intercept above. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-2cm \pm \sqrt{4\left( -{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}}{2\left( 1+{{m}^{2}} \right)} \\
& =\dfrac{-2\left( a\sqrt{1+{{m}^{2}}} \right)m \pm \sqrt{4\left( -{{a}^{2}}-{{a}^{2}}{{m}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{2}} \right)}}{2\left( 1+{{m}^{2}} \right)} \\
& =\dfrac{-am}{\sqrt{1+{{m}^{2}}}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-am}{\sqrt{1+{{m}^{2}}}} \right)+a\sqrt{1+{{m}^{2}}}=\dfrac{a}{\sqrt{1+{{m}^{2}}}}\]
So the point of contact is $\left( \dfrac{-am}{\sqrt{{{m}^{2}}+1}},\dfrac{a}{\sqrt{{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (I)(ii)(Q).
(II) The given equation of conic is
\[{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}+{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}+{{a}^{2}}{{m}^{2}}{{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
& \Rightarrow \left( 1+{{a}^{2}}{{m}^{2}} \right){{x}^{2}}+2{{a}^{2}}cmx+{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find ,
\[\begin{align}
& D={{\left( 2{{a}^{2}}cm \right)}^{2}}-4\left( 1+{{a}^{2}}{{m}^{2}} \right)\left( {{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}-{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow 4\left( -{{a}^{2}}{{c}^{2}}+{{a}^{2}}+{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow {{a}^{2}}+{{a}^{4}}{{m}^{2}}={{a}^{2}}{{c}^{2}} \\
& \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}+1 \\
& \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}+1} \\
\end{align}\]
So the equation of the tangent is
\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+1}\]
The coordinates of the point of contact of the conic are roots which we find by putting \[{{c}^{2}}={{a}^{2}}{{m}^{2}}+1,{{c}^{2}}=\sqrt{{{a}^{2}}{{m}^{2}}+1}\] from above calculation of intercept. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}{{c}^{2}} \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}\pm \sqrt{4\left( {{a}^{2}}+{{a}^{4}}{{m}^{2}}-{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}+1 \right) \right)}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}+1}}{2\left( {{a}^{2}}{{m}^{2}}+1 \right)} \\
& =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}+1}=\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}}\]
So the point of contact is $\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+1}},\dfrac{1}{\sqrt{{{a}^{2}}{{m}^{2}}+1}} \right)$. We match the columns with the obtained result and find the matching as (II)(iv)(R). (I)(ii)(Q).\[\]
(III) The given equation of conic is
\[{{y}^{2}}=4ax\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{\left( mx+c \right)}^{2}}=4ax \\
& \Rightarrow {{m}^{2}}{{x}^{2}}+{{c}^{2}}+2cmx-4ax=0 \\
& \Rightarrow {{m}^{2}}{{x}^{2}}+\left( 2cm-4a \right)x+{{c}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contact. The $x$ and $y-$coordinates of point of contact in two dimensions have to be rational numbers. So by the condition on discriminant we have,
\[\begin{align}
& D={{\left( 2cm-4a \right)}^{2}}-4{{m}^{2}}{{c}^{2}}=0 \\
& \Rightarrow 4{{c}^{2}}{{m}^{2}}+16{{a}^{2}}-16cma-4c{{m}^{2}}=0 \\
& \Rightarrow 16{{a}^{2}}-16cma=0 \\
& \Rightarrow c=\dfrac{a}{m} \\
\end{align}\]
So the equation of the tangent is
\[\begin{align}
& y=mx+\dfrac{a}{m} \\
& \Rightarrow my={{m}^{2}}x+a \\
\end{align}\]
The coordinates of the point of contact of the conic are roots which we find by putting $c=\dfrac{a}{m},{{c}^{2}}=\dfrac{{{a}^{2}}}{{{m}^{2}}}$ from calculation of intercept above. We have the $x-$coordinate of point of contact as,
\[\begin{align}
& x=\dfrac{-\left( 2cm-4a \right)\pm\sqrt{16{{a}^{2}}-16cma}}{2{{m}^{2}}} \\
& =\dfrac{-2\left( \dfrac{a}{m} \right)m+4a \pm \sqrt{16{{a}^{2}}-16\left( \dfrac{a}{m} \right)ma}}{2{{m}^{2}}} \\
& =\dfrac{a}{{{m}^{2}}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{a}{{{m}^{2}}} \right)+\dfrac{a}{m}=\dfrac{2a}{m}\]
So the point of contact is $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$. We match the columns with the obtained result and find the matching as (III)(i)(P). (II)(iv)(R). (I)(ii)(Q). .\[\]
(IV) The given equation of conic is
\[{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}\]
Let us put $y=mx+c$ in the above equation and have,
\[\begin{align}
& {{x}^{2}}-{{a}^{2}}{{\left( mx+c \right)}^{2}}={{a}^{2}} \\
& \Rightarrow {{x}^{2}}-{{a}^{2}}{{m}^{2}}{{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
&\Rightarrow \left( 1-{{a}^{2}}{{m}^{2}} \right){{x}^{2}}-2{{a}^{2}}cmx-{{a}^{2}}{{c}^{2}}-{{a}^{2}}=0 \\
\end{align}\]
The above equation is a quadratic equation whose solutions are coordinates point of contacts which have to be rational numbers. So by the condition on discriminant we find,
\[\begin{align}
& D={{\left( -2{{a}^{2}}cm \right)}^{2}}-4\left( 1-{{a}^{2}}{{m}^{2}} \right)\left( -{{a}^{2}}{{c}^{2}}-{{a}^{2}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{4}}{{c}^{2}}{{m}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{a}^{4}}{{m}^{2}}{{c}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow 4\left( {{a}^{2}}{{c}^{2}}+{{a}^{2}}-{{m}^{2}}{{a}^{4}} \right)=0 \\
& \Rightarrow {{a}^{2}}-{{a}^{4}}{{m}^{2}}=-{{a}^{2}}{{c}^{2}} \\
& \Rightarrow 1-{{a}^{2}}{{m}^{2}}=-{{c}^{2}} \\
& \Rightarrow {{c}^{2}}={{a}^{2}}{{m}^{2}}-1 \\
& \Rightarrow c=\sqrt{{{a}^{2}}{{m}^{2}}-1} \\
\end{align}\]
So the equation of the tangent is
\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-1}\]
The coordinates of the point of contact of the conic are roots which we find by putting \[{{c}^{2}}={{a}^{2}}{{m}^{2}}-1,c=\sqrt{{{a}^{2}}{{m}^{2}}-1}\] from above calculation of intercept. We have the $x-$coordinate of point of contact as,
contact as,
\[\begin{align}
& x=\dfrac{2{{a}^{2}}cm\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}{{c}^{2}} \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\
& =\dfrac{2{{a}^{2}}m\left( \sqrt{{{a}^{2}}{{m}^{2}}-1} \right)\pm \sqrt{4\left( {{a}^{2}}-{{a}^{4}}{{m}^{2}}+{{a}^{2}}\left( {{a}^{2}}{{m}^{2}}-1 \right) \right)}}{2\left( 1-{{a}^{2}}{{m}^{2}} \right)} \\
& =\dfrac{-2{{a}^{2}}m\sqrt{{{a}^{2}}{{m}^{2}}-1}}{2\left( {{a}^{2}}{{m}^{2}}-1 \right)} \\
& =\dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \\
\end{align}\]
Putting $x$ in equation of the tangent (1) we have $y-$coordinate of point of contact as,
\[y=m\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)+\sqrt{{{a}^{2}}{{m}^{2}}-1}=\dfrac{-1}{\sqrt{1+{{m}^{2}}}}\]
So the point of contact is \[\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)\] . We match the columns with the obtained result and find the matching as (IV)(iii)(S).
So the correct matches are (IV)(iii)(S), (III)(i)(P), (II)(iv)(R), (I)(ii)(Q). We are asked to choose the only INCORRECT combination in the options which is all of the options A, B,C,D.
Note: We have taken negative and positive square root for $c$ keeping the options in mind .We note that the equation ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is an equation of circle in centre-radius from, ${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of ellipse in axis form, ${{y}^{2}}=4ax$ is an equation of parabola in vertex from and ${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$ is an equation of hyperbola in axis from. We can directly find tangents and point of contact if we remember the formulae.
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