Locus of point of intersection of perpendicular tangents to ellipse frac{{{x^2}}}{9} + frac{{{y^2}}}

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10-2. Parabola, Ellipse, Hyperbola

medium

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is

${x^2} + {y^2} = 9$

${x^2} + {y^2} = 4$

${x^2} + {y^2} = 13$

${x^2} + {y^2} = 5$

Solution

(c) The locus of point of intersection of two perpendicular tangents drawn on the ellipse is ${x^2} + {y^2} = {a^2} + {b^2},$ which is called ‘director- circle’.

Given ellipse is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$,

Locus is ${x^2} + {y^2} = 13.$

Standard 11

Mathematics

Eccentricity of the ellipse $9{x^2} + 25{y^2} = 225$ is

easy

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $=\frac{x^2}{36}+\frac{y^2}{16}=1$.If $(1, \alpha)$ lies on $C$, then $10 \alpha^2$ is equal to $………$

hard

(JEE MAIN-2023)

Number of common tangents of the ellipse $\frac{{{{\left( {x – 2} \right)}^2}}}{9} + \frac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$ and the circle $x^2 + y^2 -4x + 2y + 4 = 0$ is

normal

Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle

Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .

normal

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$, then $\alpha$ is :