A tangent to a suitable conic in (Column 1) at $\left( \sqrt{3},\dfrac{1}{2} \right)$ is found to be $\sqrt{3}x+2y=4$. Then which of the following combinations is the only correct combination?
Column-1 Column-2 Column-3 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) $\left( \dfrac{-{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}-1}},\dfrac{-1}{\sqrt{{{a}^{2}}{{m}^{2}}-1}} \right)$
A. IV,(iii),S
B. II,(iv), R
C. II,(iii),R
D. IV,(iv),S
| Column-1 | Column-2 | Column-3 | |||
| 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: To solve this problem, we should consider the combinations given in the question option by option and decide which option suits the answer. We should get the value of a by substituting the given point $\left( \sqrt{3},\dfrac{1}{2} \right)$ in the curves in column-1. After getting the value of a, we should use the equation of tangent in the column-2. When two lines $ax+by+c=0$ is same as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$, then the relation between the co-efficients is $\dfrac{a}{{{a}_{1}}}=\dfrac{b}{{{b}_{1}}}=\dfrac{c}{{{c}_{1}}}$. Using the relation $\dfrac{a}{{{a}_{1}}}=\dfrac{b}{{{b}_{1}}}$, we can find the value of the slope m. We should check if the condition holds good for $\dfrac{c}{{{c}_{1}}}$using the calculated values of a and m.
Complete step by step answer:
We are given a point on the curve as $A\left( \sqrt{3},\dfrac{1}{2} \right)$ and the tangent to the point A as $\sqrt{3}x+2y=4$. If we observe the options, we can see that the curve IV in the column-1 is matched to the point S in column-3 and the curve II in the column-1 is matched to the point R in column-3. The only changing parameter is the tangent to the curve. So, we need not check for the column-3 in this question. So, by taking option by option, we will substitute the point in the curve and get the value of a.
Let us consider the curve IV in the column-1
${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$. Substituting $A\left( \sqrt{3},\dfrac{1}{2} \right)$ in this equation, we get
$\begin{align}
& {{\left( \sqrt{3} \right)}^{2}}-{{a}^{2}}{{\left( \dfrac{1}{2} \right)}^{2}}={{a}^{2}} \\
& 3-\dfrac{{{a}^{2}}}{4}={{a}^{2}} \\
& 3=\dfrac{5}{4}{{a}^{2}} \\
& {{a}^{2}}=\dfrac{12}{5} \\
\end{align}$
Let us consider option-A. The tangent is given by the row (iii) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{\dfrac{12}{5}{{m}^{2}}-1} \\
& mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}-1}=0 \\
\end{align}$
When two lines $ax+by+c=0$ is same as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$, then the relation between the co-efficients is $\dfrac{a}{{{a}_{1}}}=\dfrac{b}{{{b}_{1}}}=\dfrac{c}{{{c}_{1}}}$.
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. By the above relation, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{m}^{2}}-1}}{-4}$
From the first two terms, we get the value of m as
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this value of m in the above equation, we get
$\begin{align}
& \dfrac{\dfrac{-\sqrt{3}}{2}}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{\left( \dfrac{-\sqrt{3}}{2} \right)}^{2}}-1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}\times \dfrac{3}{4}-1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{36-20}{20}}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-1}{\sqrt{20}} \\
\end{align}$
By the first option, we did not get the correct inequality, which is why this is not the correct answer.
Let us consider option-D with the calculated value of ${{a}^{2}}=\dfrac{12}{5}$ as the curve is same.
The tangent is given by the row (iv) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{\dfrac{12}{5}{{m}^{2}}+1} \\
& mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}+1}=0 \\
\end{align}$
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. Using the similar lines concept, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{m}^{2}}+1}}{-4}$
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this, we get
$\begin{align}
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}\times \dfrac{3}{4}+1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{36+20}{20}}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-\sqrt{14}}{2\sqrt{20}} \\
\end{align}$
By the fourth option, we did not get the correct inequality, which is why this is not the correct answer.
Let us consider option-B.
Let us consider the curve II in the column-1
${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$. Substituting $A\left( \sqrt{3},\dfrac{1}{2} \right)$ in this equation, we get
$\begin{align}
& {{\left( \sqrt{3} \right)}^{2}}+{{a}^{2}}{{\left( \dfrac{1}{2} \right)}^{2}}={{a}^{2}} \\
& 3+\dfrac{{{a}^{2}}}{4}={{a}^{2}} \\
& 3=\dfrac{3}{4}{{a}^{2}} \\
& {{a}^{2}}=4 \\
\end{align}$
The tangent is given by the row (iv) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{4{{m}^{2}}+1} \\
& mx-y+\sqrt{4{{m}^{2}}+1}=0 \\
\end{align}$
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. Using the similar lines concept, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{4{{m}^{2}}+1}}{-4}$
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this, we get
$\begin{align}
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{4\times \dfrac{3}{4}+1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{4}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{2}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-1}{2} \\
\end{align}$
So, we got the correct inequality with option-B.
$\therefore $The correct combination for the given point and tangent is II,(iv), R. The answer is option-B.
Note:
Students try to solve the equation containing m which we got by substituting the point in the equation of tangent. For example, if we consider option-A, they try to solve for m in $mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}-1}=0$ by substituting the point. After getting the values of m, they try to use them in the column-3 to check. This procedure will lead to the answer but it is a process that involves more calculations. We will not use the tangent equation given in the question in this process. So, we should always think about a procedure with less few of calculations and we should use all the information given in the question.
Complete step by step answer:
We are given a point on the curve as $A\left( \sqrt{3},\dfrac{1}{2} \right)$ and the tangent to the point A as $\sqrt{3}x+2y=4$. If we observe the options, we can see that the curve IV in the column-1 is matched to the point S in column-3 and the curve II in the column-1 is matched to the point R in column-3. The only changing parameter is the tangent to the curve. So, we need not check for the column-3 in this question. So, by taking option by option, we will substitute the point in the curve and get the value of a.
Let us consider the curve IV in the column-1
${{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}$. Substituting $A\left( \sqrt{3},\dfrac{1}{2} \right)$ in this equation, we get
$\begin{align}
& {{\left( \sqrt{3} \right)}^{2}}-{{a}^{2}}{{\left( \dfrac{1}{2} \right)}^{2}}={{a}^{2}} \\
& 3-\dfrac{{{a}^{2}}}{4}={{a}^{2}} \\
& 3=\dfrac{5}{4}{{a}^{2}} \\
& {{a}^{2}}=\dfrac{12}{5} \\
\end{align}$
Let us consider option-A. The tangent is given by the row (iii) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{\dfrac{12}{5}{{m}^{2}}-1} \\
& mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}-1}=0 \\
\end{align}$
When two lines $ax+by+c=0$ is same as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$, then the relation between the co-efficients is $\dfrac{a}{{{a}_{1}}}=\dfrac{b}{{{b}_{1}}}=\dfrac{c}{{{c}_{1}}}$.
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. By the above relation, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{m}^{2}}-1}}{-4}$
From the first two terms, we get the value of m as
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this value of m in the above equation, we get
$\begin{align}
& \dfrac{\dfrac{-\sqrt{3}}{2}}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{\left( \dfrac{-\sqrt{3}}{2} \right)}^{2}}-1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}\times \dfrac{3}{4}-1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{36-20}{20}}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-1}{\sqrt{20}} \\
\end{align}$
By the first option, we did not get the correct inequality, which is why this is not the correct answer.
Let us consider option-D with the calculated value of ${{a}^{2}}=\dfrac{12}{5}$ as the curve is same.
The tangent is given by the row (iv) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{\dfrac{12}{5}{{m}^{2}}+1} \\
& mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}+1}=0 \\
\end{align}$
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. Using the similar lines concept, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}{{m}^{2}}+1}}{-4}$
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this, we get
$\begin{align}
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{12}{5}\times \dfrac{3}{4}+1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{\dfrac{36+20}{20}}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-\sqrt{14}}{2\sqrt{20}} \\
\end{align}$
By the fourth option, we did not get the correct inequality, which is why this is not the correct answer.
Let us consider option-B.
Let us consider the curve II in the column-1
${{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}$. Substituting $A\left( \sqrt{3},\dfrac{1}{2} \right)$ in this equation, we get
$\begin{align}
& {{\left( \sqrt{3} \right)}^{2}}+{{a}^{2}}{{\left( \dfrac{1}{2} \right)}^{2}}={{a}^{2}} \\
& 3+\dfrac{{{a}^{2}}}{4}={{a}^{2}} \\
& 3=\dfrac{3}{4}{{a}^{2}} \\
& {{a}^{2}}=4 \\
\end{align}$
The tangent is given by the row (iv) in column-2. By using the value of ${{a}^{2}}$ in the tangent equation, we get
$\begin{align}
& y=mx+\sqrt{4{{m}^{2}}+1} \\
& mx-y+\sqrt{4{{m}^{2}}+1}=0 \\
\end{align}$
The tangent equation given in the question is $\sqrt{3}x+2y-4=0$. Using the similar lines concept, we can write that
$\dfrac{m}{\sqrt{3}}=\dfrac{-1}{2}=\dfrac{\sqrt{4{{m}^{2}}+1}}{-4}$
$\begin{align}
& \dfrac{m}{\sqrt{3}}=\dfrac{-1}{2} \\
& m=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
Using this, we get
$\begin{align}
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{4\times \dfrac{3}{4}+1}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{\sqrt{4}}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{2}{-4} \\
& \dfrac{-1}{2}=\dfrac{-1}{2}=\dfrac{-1}{2} \\
\end{align}$
So, we got the correct inequality with option-B.
$\therefore $The correct combination for the given point and tangent is II,(iv), R. The answer is option-B.
Note:
Students try to solve the equation containing m which we got by substituting the point in the equation of tangent. For example, if we consider option-A, they try to solve for m in $mx-y+\sqrt{\dfrac{12}{5}{{m}^{2}}-1}=0$ by substituting the point. After getting the values of m, they try to use them in the column-3 to check. This procedure will lead to the answer but it is a process that involves more calculations. We will not use the tangent equation given in the question in this process. So, we should always think about a procedure with less few of calculations and we should use all the information given in the question.
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