JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the tangents on the ellipse \(4x^2 + y^2 = 8\) at the points \((1, 2)\) and \((a, b)\) are perpendicular to each other, then \(a^2\) is equal to
- A \(\frac{2}{{17}}\)
- B \(\frac{4}{{17}}\)
- C \(\frac{64}{{17}}\)
- D \(\frac{128}{{17}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{{17}}\)
Step-by-step Solution
Detailed explanation
\(4{a^2} + {b^2} = 8\,\,\,\,\,\,.......\left( 1 \right)\) \({\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = - \frac{{4x}}{y} = - 2\) \( \Rightarrow - \frac{{4a}}{b} = \frac{1}{2}\) \(b = - 8a\) \( \Rightarrow {b^2} = 64{a^2}\) \(68{a^2} = 8,{a^2} = \frac{2}{{17}}\)
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