JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the normal to the ellipse \(3x^2 + 4y^2 = 12\) at a point \(P\) on it is parallel to the line, \(2x + y = 4\) and the tangent to the ellipse at \(P\) passes through \(Q(4, 4)\) then \(PQ\) is equal to
- A \(\frac{{\sqrt {157} }}{2}\)
- B \(\frac{{5\sqrt 5 }}{2}\)
- C \(\frac{{\sqrt {221} }}{2}\)
- D \(\frac{{\sqrt {61} }}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{{5\sqrt 5 }}{2}\)
Step-by-step Solution
Detailed explanation
\(3{x^2} + 4{y^2} = 12\) \(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1\) \(x = 2\,\cos \theta ,y = \sqrt 3 \sin \theta \) Equation of noraml is \(\frac{{{a^2}x}}{{{x_1}}} + \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}\)…
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