Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=m x+c$ is a common tangent to the parabola $y^2=4 \sqrt{k} x$ and the circle $2 x^2+2 y^2=k$ then the product of the slopes of such common tangents is
Options:
Solution:
2255 Upvotes
Verified Answer
The correct answer is:
$-1$
The equation of a tangent to $y^2=4 \sqrt{k} x$ is
$y=m x+\sqrt{k} / m$, where $m$ is the slope of tangent.
If it touches the $2 x^2+2 y^2=k$, then
$\begin{aligned} & \left|\frac{\sqrt{k} / m}{\sqrt{1+m^2}}\right|=\sqrt{\frac{k}{2}} \\ \Rightarrow \quad m \sqrt{1+m^2} & =\sqrt{2} \Rightarrow m^4+m^2-2=0 \\ \Rightarrow \quad\left(m^2+2\right)\left(m^2-1\right) & =0 \Rightarrow m= \pm 1\end{aligned}$
Substituting these values in $y=m x+\frac{\sqrt{k}}{m}$, the equation of common tangents are $y=x+\sqrt{k}$ and
$\begin{aligned} & y=-x-\sqrt{k} \\ & \therefore \quad m_1=1, m_2=-1 \\ & \text { Now, } m_1 m_2=1 \times-1=-1\end{aligned}$
$y=m x+\sqrt{k} / m$, where $m$ is the slope of tangent.
If it touches the $2 x^2+2 y^2=k$, then
$\begin{aligned} & \left|\frac{\sqrt{k} / m}{\sqrt{1+m^2}}\right|=\sqrt{\frac{k}{2}} \\ \Rightarrow \quad m \sqrt{1+m^2} & =\sqrt{2} \Rightarrow m^4+m^2-2=0 \\ \Rightarrow \quad\left(m^2+2\right)\left(m^2-1\right) & =0 \Rightarrow m= \pm 1\end{aligned}$
Substituting these values in $y=m x+\frac{\sqrt{k}}{m}$, the equation of common tangents are $y=x+\sqrt{k}$ and
$\begin{aligned} & y=-x-\sqrt{k} \\ & \therefore \quad m_1=1, m_2=-1 \\ & \text { Now, } m_1 m_2=1 \times-1=-1\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.