If the tangent at the point $(1,2)$ on the ellipse $3 x^2+4 y^2=19$ is also a tangent to the parabola $y^2-k x=0$, then $k=$

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Aicharya · Accepted Answer

Equation of tangent at the point $(1,2)$ on the ellipse $3 x^2+4 y^2=19$ is $ 3 \cdot x \cdot 1+4 \cdot y \cdot 2=19 $

This is also tangent to the parabola

From Eqs. (i) and (ii), we get $ y^2-k\left(\frac{19-8 y}{3}\right)=0 $ $ \begin{array}{lrl} \Rightarrow & 3 y^2+8 k y-19 k=0 \text { has equal roots, so } \\ \therefore & D=(8 k)^2-4(3)(-19 k)=0 \\ \Rightarrow & 64 k^2=-4 \cdot 3 \cdot 19 k \\ \Rightarrow & k=\frac{-57}{16} . \end{array} $

If the tangent at the point $(1,2)$ on the ellipse $3 x^2+4 y^2=19$ is also a tangent to the parabola $y^2-k…

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