Question

If the tangents on the ellipse $4x^2+y^2=8$ at the points $(1,2)\mathrm{and}(a,b)$ are perpendicular to each other, then $a^2$ is equal to:

• A. $\frac{2}{17}$
• B. $\frac{64}{17}$
• C. $\frac{128}{17}$
• D. $\frac{4}{17}$

Answer 1

The correct option is A

$\frac{2}{17}$

Explanation for the correct option.

Step 1: Find the slope of the tangent.

The equation of ellipse is given to be $4x^2+y^2=8$.

Now, to find the equation of the tangent, we will need the slope for which will we differentiate the given equation with respect to $x$ and it will be,

$\begin{array}{rcl}8x+2y\frac{dy}{dx}& =& 0\\ \Rightarrow \frac{dy}{dx}& =& \frac{-8x}{2y}\\ & =& \frac{-4x}{y}\end{array}$

Step 2: Find the value of $\frac{dy}{dx}$ for the given points.

$\begin{array}{rcl}{\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}& =& \frac{-4\left(1\right)}{2}\\ & =& -2\\ {\left(\frac{dy}{dx}\right)}_{\left(a,b\right)}& =& \frac{-4\left(a\right)}{b}\\ & =& \frac{-4a}{b}\end{array}$

Step 3: Find the value of $a^2$.

It is given that the tangents are perpendicular to each other, so the product of the slopes will be $-1$. So,

$\begin{array}{rcl}\left(-2\right)\times \frac{-4a}{b}& =& -1\\ \Rightarrow 8a& =& -b\end{array}$

Now, as the point $\left(a,b\right)$ is on ellipse.

So it will satisfy the given equation of ellipse. That means

$\begin{array}{rcl}4a^2+b^2& =& 8\\ & \Rightarrow & 4a^2+{\left(-8a\right)}^2=8\left(\mathrm{by}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}b\right)\\ \Rightarrow 68a^2& =& 8\\ \Rightarrow a^2& =& \frac{2}{17}\end{array}$

Hence, option (A) is correct.