# Latus Rectum of Hyperbola

## Trending Questions

**Q.**The point of intersection of two tangents to the hyperbola x2a2−y2b2=1, the product of whose slopes is c2, lies on the curve.

**Q.**

For some$\theta \in (0,\frac{\mathrm{\pi}}{2})$, if the eccentricity of the hyperbola, ${x}^{2}-{y}^{2}{\mathrm{sec}}^{2}\left(\theta \right)=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, ${x}^{2}{\mathrm{sec}}^{2}\left(\theta \right)+{y}^{2}=5$, then the length of the latus rectum of the ellipse, is:

$\frac{4\sqrt{5}}{3}$

$\frac{2\sqrt{5}}{3}$

$2\sqrt{6}$

$\sqrt{30}$

**Q.**The eccentricity of the hyperbola −x2a2+y2b2=1 is

**Q.**Let P be any point on the hyperbola x2a2−y2b2=1 such that the absolute difference of the distances of P from the two foci is 12. If the eccentricity of the hyperbola is 2, then the length of the latus rectum is

- 4√3 unit
- 18 unit
- 2√3 unit
- 36 unit

**Q.**

The equation of the hyperbola having its eccentricity $2$ and the distance between its foci is $8$, is:

$\frac{{x}^{2}}{12}-\frac{{y}^{2}}{4}=1$

$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{12}=1$

$\frac{{x}^{2}}{8}-\frac{{y}^{2}}{2}=1$

$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$

**Q.**Equation of one of the latus rectum of the hyperbola (10x−5)2+(10y−2)2=9(3x+4y−7)2 is

- 30x+40y−23=0
- 30x+40y+23=0
- 40x+30y−23=0
- 40x+30y+23=0

**Q.**The area of the figure bounded by the parabolas x=−2y2 and x=1−3y2 is

- 43 square units
- 23 square units
- 37 square units
- 67 square units

**Q.**The equation of the latus rectum of the parabola represented by equation y2+2Ax+2By+C=0 is

**Q.**

The latus-rectum of the hyperbola

16x2−9y2=144 is

16/3

32/3

8/3

4/3

**Q.**A is the vertex of the hyperbola x2−2y2−2√5x−4√2y−3=0, B is one of the end points of latus rectum and C is the focus of the hyperbola. If A, B and C lies on same side of conjugate axis, then the area of the triangle ABC is

- 2 sq. units
- √32−1 sq. units
- 1−√23 sq. units
- √32+1 sq. units

**Q.**If the coordinates of the vertices of a hyperbola are (9, 2) and (1, 2) and the distance between its foci is 10 units. Then

- length of its latus rectum is 9 units
- length of its latus rectum is 92 units
- equation of the hyperbola is (x−5)29- \dfrac{(y-2)^2}{16}=1$
- equation of the hyperbola is (x−5)216- \dfrac{(y-2)^2}{9}=1$

**Q.**Find the equation of ellipse whose eccentricity is 23, latus rectum is 5 and the centre is (0, 0).

**Q.**An ellipse intersects the hyperbola 2x2−2y2=1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

- the foci of ellipse are (±√2, 0)
- Equation of ellipse is x2+2y2=1
- the foci of ellipse are (±1, 0)
- equation of ellipse are x2+2y2=4

**Q.**Let P be any point on the hyperbola x2a2−y2b2=1 such that the absolute difference of the distances of P from the two foci is 12. If the eccentricity of the hyperbola is 2, then the length of the latus rectum is

- 4√3 unit
- 18 unit
- 2√3 unit
- 36 unit

**Q.**The locus of the point of interaction of two tangents of the hyperbola x2a2−y2b2=1 which make an angle α with one another is

**Q.**

If a hyperbola has a length of its conjugate axis equal to $5$ and the distance between its foci is $13$, then the eccentricity of the hyperbola is:

$9$

$\frac{13}{12}$

$\frac{32}{3}$

$\frac{64}{3}$

**Q.**If focii of x2a2−y2b2=1 coincide with the focii of x225+y29=1 and eccentricity of the hyperbola is 2, then

- a2+b2=14
- There is a director circle of the hyperbola
- Length of latus rectum of the hyperbola is 12
- Centre of the director circle is (0, 0)

**Q.**

Write the length of the latus-rectum of the hyperbola16x2−9y2=144

**Q.**Let a+b+c=10, a, b, c>0. Then the maximum value of a2b3c5 is

- (15)3×100
- (15)3×10
- (15)3×1000
- (60)3

**Q.**The solution of dydx=x2+y2+12xy satisfying y(1) = 0 is

- an ellipse with e =1√2
- a hyperbola with e = √2
- a circle whose radius is 1
- a curve symmetric about origin

**Q.**A variable plane forms a tetrahedron of constant volume 64K3 with the coordinate planes and the origin. The locus of the centroid of the tetrahedron is

- xyz=6k3
- x3+y3+z3=6K2
- x2+y2+z2=4K2
- x−2+y−2+z−2=4k−2

**Q.**A plane intersects the co ordinate axes at A, B, C. If O=(0, 0, 0) and (1, 1, 1) is the centroid of the tetrahedron OABC, then the sum of the reciprocals of the intercepts of the plane

- 12
- 43
- 1
- 34

**Q.**The hyperbola x2a2−y2b2=1 has its conjugate axis of length 5 and passes through the point (2, 1). The length of latus rectum is :

- 54√29
- 58√29
- √294
- √29

**Q.**Find the vertex, axis, focus, directrix and latus rectum of the parabola, also draw their rough sketches

4y2+12x−20y+67=0

**Q.**

The area (in square units) of the quadrilateral formed by the two pairs of lines

l2x2−m2y2−n(lx+my)=0

and l2x2−m2y2+n(lx−my)=0 is

**Q.**Let the eccentricity of the hyperbola x2a2−y2b2=1 be reciprocal to that of the ellipse x2+4y2=4. lf the hyperbola passes through a focus of the ellipse, then

- the equation of the hyperbola is x23−y22=1
- a focus of the hyperbola is (2, 0)
- the equation of the hyperbola is x2−3y2=3
- the eccentricity of the hyperbola is √53

**Q.**Equation of one of the latus rectum of the hyperbola (10x−5)2+(10y−2)2=9(3x+4y−7)2 is

- 30x+40y−23=0
- 40x+30y−23=0
- 30x+40y+23=0
- 40x+30y+23=0

**Q.**If the coordinates of the vertices of a hyperbola are (9, 2) and (1, 2) and the distance between its foci is 10 units. Then

**Q.**

If the latus rectum of an hyperbola be 8 and eccentricity be 3√5 , then the equation of the hyperbola is

4x2−5y2=100

5x2−4y2=100

4x2+5y2=100

5x2+4y2=100

**Q.**The eccentricity of the hyperbola whose latus-return is 8 and length of the conjugate axis is equal to half the distance between the foci, is

- 43
- 4√3
- 2√3
- None of these