JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
In the line \( \alpha x+4y=\sqrt{7} \), where \( \alpha\in R \) touches the ellipse \( 3x^{2}+4y^{2}=1 \) at the point P in the first quadrant, then one of the focal distances of P is :
- A \( \frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{11}} \)
- B \( \frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{5}} \)
- C \( \frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{5}} \)
- D \( \frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}} \)
Answer & Solution
Correct Answer
(D) \( \frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}} \)
Step-by-step Solution
Detailed explanation
\( \alpha x+4y-\sqrt{7}=0 \) touches \( 3x^{2}+4y^{2}=1 \) \( \therefore c^{2}=a^{2}m^{2}+b^{2} \) \( \frac{7}{16}=\frac{1}{3}\times\frac{\alpha^{2}}{16}+\frac{1}{4}\Rightarrow\alpha=3,-3 \) Tangent is \( 3x+4y-\sqrt{7}=0 \) Let the point of contact is \( P(x_{1}y_{1}) \)…
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