If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve:

\(\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1\)

\(\frac{1}{{4{x^2}}} + \frac{1}{{2{y^2}}} = 1\)

\(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1\)

\(\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1\)

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the

1.	If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve:
A.	\(\frac{1}{{4{x^2}}} + \frac{1}{{2{y^2}}} = 1\)
B.	\(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1\)
C.	\(\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1\)
D.	\(\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1\)
Answer» D. \(\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1\)

Discussion

No Comment Found

Related MCQs

Let P (4, -4) and Q (9, 6) be two points on the parabola, y2 = 4x and let X be any point on the POQ of this parabola, where O is the vertex of this parabola, such that the area of ΔPXQ is maximum. Then, this maximum area (in sq units) is
Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?
If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x - a2) = 0 and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of ‘a’ is:
If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve:
If the foci of the ellipse \(b^2x^2 +16y^2 = 16b^2\) and the hyperbola \(81x^2 -144y^2 = \dfrac{81 \times 144}{25}\) coincide, then the value of b, is
A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Find the eccentricity of the hyperbola.
Let \(0 < {\rm{\theta }} < \frac{{\rm{\pi }}}{2}\). If the eccentricity of the hyperbola \(\frac{{{{\rm{x}}^2}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{\theta }}}} - \frac{{{{\rm{y}}^2}}}{{{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}} = 1\) is greater than 2, then the length of its latus rectum lies in the interval:
If the area (in sq. units) of the region {(x, y) : y2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0} is \({\rm{a}}\sqrt 2 + {\rm{b}}\), then a – b is equal to:
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at \(\left( {0,\;5\sqrt 3 } \right),\) then the length of its latus rectum is:
If the vertices of a hyperbola be at (-2, 0) and (2, 0) and one of its foci be at (-3, 0), then which one of the following points does not lie on this hyperbola?

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve:

Discussion

No Comment Found

Related MCQs

Reply to Comment