The focal chord of the parabola y^2=32x touches the ellipse (x^2)/(4^2)+(y^2)/(2^2)=1 in the first quadrant at the point

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) 2,√3. Let the tangent to the ellipse at 4cos⁡θ,2sin⁡θ be (xcos⁡θ)/4+(ysin⁡θ)/2=1 ∵ it passes through the focus 8,0 ⇒cos⁡θ=1/2⇒θ=(π)/3 ⇒ point is 4/2,(2√3)/2≡2,√3

The focal chord of the parabola $y^{2} = 32 x$ touches the ellipse $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} = 1$ in the first quadrant at the point

Solution

More Ellipse Questions

The focal chord of the parabola y2=32x touches the ellipse x242+y222=1 in the first quadrant at the point

Solution

More Ellipse Questions

The focal chord of the parabola $y^{2} = 32 x$ touches the ellipse $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} = 1$ in the first quadrant at the point