Question

Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, – 4), one focus at (4, – 4) and one vertex at (5, – 4). If mx – y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m² is equal to______.

Answer 1

Correct answer is 3

Given C(3,-4), S(4,-4)

and A(5,–4) 

Hence, a = 2 & ae = 1

\(\Rightarrow\) e = \(\frac{1}{2}\)

\(\Rightarrow\) b² = 3

So, E: \(\frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1\)

Intersecting with given tangent.

\(\frac{x^2 + 6x + 9}{4} + \frac{m^2x^2}{3} = 1\)

Now, D = 0 (as it is tangent)

 So, 5m² = 3.