JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the point \(P\) of the focal chord \(P Q\) of the parabola \(y^2=16 x\) be \((1,-4)\). If the focus of the parabola divides the chord PQ in the ratio \(\mathrm{m}: \mathrm{n}\), \(\operatorname{gcd}(m, n)=1\), then \(m^2+n^2\) is equal to :
- A \(17\)
- B \(10\)
- C \(37\)
- D \(26\)
Answer & Solution
Correct Answer
(A) \(17\)
Step-by-step Solution
Detailed explanation
\(y^2=16 x ; a=4 \quad\) focus \(S \equiv(4,0)\) \(\begin{aligned} & 2 \mathrm{at}_1=-4 \\ & \Rightarrow 2(4) \mathrm{t}_1=-4 \\ & \Rightarrow \mathrm{t}_1=-\frac{1}{2} \\ & \because \mathrm{t}_1 \mathrm{t}_2=-1\end{aligned}\)…
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