If $5 x-2 y+k=0$ is a tangent to the parabola $y^2=6 x$, then their point of contact is

Question

Aicharya · Accepted Answer

Given curve $ y^2=6 x $ difference w.r.t ' $x$ ' $ \Rightarrow \quad 2 y \frac{d y}{d x}=6 $

Given equation of tangent is $5 x-2 y+k=0$ $ \begin{aligned} \quad \text { Slope } & =\frac{5}{2} \\ \therefore \quad \frac{3}{y} & =\frac{5}{2} \Rightarrow y=\frac{6}{5} \end{aligned} $ Substitute value of $y$ in $y^2=6 x$ $ \begin{aligned} \left(\frac{6}{5}\right)^2 & =6 x \Rightarrow \frac{36}{25}=6 x \\ \Rightarrow \quad x & =\frac{6}{25} \end{aligned} $ Hence, point of contact is $\left(\frac{6}{25}, \frac{6}{5}\right)$

If $5 x-2 y+k=0$ is a tangent to the parabola $y^2=6 x$, then their point of contact is

Solution