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 $x^2 + y^2 = (a + b)^2$ .
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 $x^2 + y^2 = (a + b)^2$ .
- A
Statement $-1$ is true, statement $-2$ is true but statement $-1$ is not the correct explanation for statement $-2$
- B
Statement $-1$ is true, statement $-2$ is false
- C
Statement $-1$ is false, statement $-2$ is true
- D
Both statements are true, and statement $-1$ is the true explanation of statement $-2$
Similar Questions
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$ . If $L$ is also a tangent to the ellipse $\frac{ x ^{2}}{2}+\frac{ y ^{2}}{ b }=1,$ then the value of $b$ is equal to ..... .
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$ . If $L$ is also a tangent to the ellipse $\frac{ x ^{2}}{2}+\frac{ y ^{2}}{ b }=1,$ then the value of $b$ is equal to ..... .
- [JEE MAIN 2021]
The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
If end points of latus rectum of an ellipse are vertices of a square, then eccentricity of ellipse will be -
If end points of latus rectum of an ellipse are vertices of a square, then eccentricity of ellipse will be -
Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ is
Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ is
Length of common chord of the ellipse ${\frac{{\left( {x - 2} \right)}}{9}^2} + {\frac{{\left( {y + 2} \right)}}{4}^2} = 1$ and the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$
Length of common chord of the ellipse ${\frac{{\left( {x - 2} \right)}}{9}^2} + {\frac{{\left( {y + 2} \right)}}{4}^2} = 1$ and the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$