JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, \(\frac{x^{2}}{4}+\frac{y^{2}}{2}=1\) from any of its foci?
- A \((-1, \sqrt{3})\)
- B \((-1, \sqrt{2})\)
- C \((-2, \sqrt{3})\)
- D \((1,2)\)
Answer & Solution
Correct Answer
(A) \((-1, \sqrt{3})\)
Step-by-step Solution
Detailed explanation
Let foot of perpendicular is \(( h , k )\) \(\frac{x^{2}}{4}+\frac{y^{2}}{2}=1 \quad(\) Given \()\) \(a=2, b=\sqrt{2}, e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}\) \(\therefore\) Focus \(( ae , 0)=(\sqrt{2}, 0)\) Equation of tangent \(y=m x+\sqrt{a^{2} m^{2}+b^{2}}\)…
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