AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Equation of tangent at point (1,2) at the ellipse 3x2+4y2=19 is
3x(1)+4y(2)=19
3x+8y=19
This line is also tangent to the parabola
y2−kx=0
Substitute the value in the above equation.
y2−k(

19−8y

3

)=0
3y2+8ky−19k=0
The equation has equal roots.
(8k)2−4(3)(−19k)=0
64k2=−(12×19)k
k=

−57

16