Question

If the tangent to the conic, $y-6={{x}^{2}}$ at (2, 10) touches the circle ${{x}^{2}}+{{y}^{2}}+8x-2y=k$ (for some fixed k) at a point $\left( \alpha ,\beta \right)$ ; then $\left( \alpha ,\beta \right)$ is ;
A.$\left( -\dfrac{4}{17},\dfrac{1}{17} \right)$
B. $\left( -\dfrac{7}{17},\dfrac{6}{17} \right)$
C. $\left( -\dfrac{6}{17},\dfrac{10}{17} \right)$
D.$\left( -\dfrac{8}{17},\dfrac{2}{17} \right)$

Answer 1

Hint: Differentiate the conic section with respect to x and find $\left( \dfrac{dy}{dx} \right)$ at point (2, 10) then find the equation of tangent using formula $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ where m is slope and $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point on the line. Then write circle equation in form of ${{\left( x+y \right)}^{2}}+{{\left( y+f \right)}^{2}}=c$ then find centre and radius of circle and further find the value of $\alpha ,\beta $ .

Complete step-by-step answer:

 We are given the equation of a conic section which is $y-6={{x}^{2}}$ .

If we have to find the slope of tangent of the any conic section at appoint let’s say $\left( {{x}_{1}},{{y}_{1}} \right)$ then we have to differentiate it with respect to x and find $\dfrac{dy}{dx}$ . Then substitute points $\left( {{x}_{1}},{{y}_{1}} \right)$ in the place of (x, y) to get the results. So, the equations of curve is,

$y-6={{x}^{2}}$

On differentiating we get,

$\dfrac{dy}{dx}=2x$

Using formula, $\dfrac{d}{dx}\left( y \right)=\dfrac{dy}{dx},\dfrac{d\left( \text{constant} \right)}{dx}=0$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$

So at point (2, 10) the slope is

$\dfrac{dy}{dx}=2\times 2=4$

Now as we know the slope of line and the point it is passing so we can get the equation of line

$\left( {{x}_{1}}-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$

Where $\left( {{x}_{1}}-{{y}_{1}} \right)$ is the point and $m=\dfrac{dy}{dx}$ which is slope.
So the equation of line is,

Y – 10 = 4 (x – 2)

On simplifying we get,

y – 10 = 4x – 8

Now adding ‘10’ to both the sides of equation we get,

y = 4x + 2

As we know the equation of circle is,

${{x}^{2}}+{{y}^{2}}+8x-2y=k$

Then we can rearrange it and write it as,

${{x}^{2}}+8x+16+{{y}^{2}}-2y+1=k+17$

So, ${{\left( x+4 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=k+17$

So the centre of circle is (-4, 1) which can be found out by using the fact that if the equation of circle is ${{\left( x+y \right)}^{2}}+{{\left( y+f \right)}^{2}}=c$ the radius is equal to (-y, -f).
Now as we know that a common target of the circle is y = 4x + 2 which can also be written as 4x – y + 2 = 0.

We can find out the distance between the line and centre of circle to represent it as radius of the circle using formula if the point is (e, f) and the equation of line is ax + by + c = 0 then the distance is, $\dfrac{\left| ae+bf+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$

So the point is (-4, 1) and the line is 4x – y + 2 = 0 then the radius is,

$r=\dfrac{\left| 4x\left( -4 \right)+\left( -1 \right)\times 1+2 \right|}{\sqrt{{{\left( -4 \right)}^{2}}+{{1}^{2}}}}$

Hence $r=\dfrac{15}{\sqrt{17}}$

As we already know of radius of circle by equation of circle is $\sqrt{k+17}$ then we can equate it to find out,

$\sqrt{k+17}=\dfrac{15}{\sqrt{17}}$

Now by squaring both sides we get,

$k+17=\dfrac{225}{17}$

So substituting k + 17 value in the equation of circle we get,

${{\left( x+4 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=\dfrac{225}{17}$

 As the point $\left( \alpha ,\beta \right)$ satisfy the circle we can say that,

${{\left( \alpha +4 \right)}^{2}}+{{\left( \beta -1 \right)}^{2}}=\dfrac{225}{17}$

$\left( \alpha ,\beta \right)$also satisfy the equation of tangent then we can say that,

$\beta =4\alpha +2$

So now substituting $\beta =4\alpha +2$ in equation of circle satisfy $\left( \alpha ,\beta

\right)$ we get,

${{\left( \alpha +4 \right)}^{2}}+{{\left( 4\alpha +1 \right)}^{2}}=\dfrac{225}{17}$

Now by breaking and expanding using the formula ${{\left( x+y

\right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$ we get,

$\begin{align}

  & {{\alpha }^{2}}+8\alpha +16+16{{\alpha }^{2}}+8\alpha +1=\dfrac{225}{17} \\

 & \Rightarrow 17{{\alpha }^{2}}+16\alpha +17=\dfrac{225}{17} \\

\end{align}$

Now by doing cross multiplication we get,

$289{{\alpha }^{2}}+272\alpha +289=225$

By subtracting 225 to both the sides we get,

$289{{\alpha }^{2}}+272\alpha +64=0$

We can represent it as,

${{\left( 17\alpha \right)}^{2}}+2\times \left( 17\alpha \right)\times 8+{{\left( 8 \right)}^{2}}=0$

So now we can write,

${{x}^{2}}+2xy+{{y}^{2}}={{\left( x+y \right)}^{2}}$

Hence the equation can be written as,

${{\left( 17\alpha +8 \right)}^{2}}=0$

Hence $\alpha =-\dfrac{8}{17}$

As we know $\beta =4\alpha +2$ so substituting $\alpha =-\dfrac{8}{17}$ we get,

$\begin{align}

  & \beta =4\times \left( -\dfrac{8}{17} \right)+2 \\

 & =-\dfrac{32}{17}+2=\dfrac{2}{17} \\

\end{align}$

Hence $\beta =\dfrac{2}{17}$

So the correct option is ‘D’.


Note: Students generally forget to use the formula that the distance between centre and tangent of any point of the circle is represented as radius of circle. Only after taking the value of $\alpha $ will it give an answer, no need to find $\beta $.