1
IIT-JEE 1999
MCQ (Single Correct Answer)
+2
-0.5
Let $$P$$ $$\left( {a\,\sec \,\theta ,\,\,b\,\tan \theta } \right)$$ and $$Q$$ $$\left( {a\,\sec \,\,\phi ,\,\,b\,\tan \,\phi } \right)$$, where $$\theta + \phi = \pi /2,$$, be two points on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$.

If $$(h, k)$$ is the point of intersection of the normals at $$P$$ and $$Q$$, then $$k$$ is equal to

A
$${{{a^2} + {b^2}} \over a}$$
B
$$ - \left( {{{{a^2} + {b^2}} \over a}} \right)$$
C
$${{{a^2} + {b^2}} \over b}$$
D
$$ - \left( {{{{a^2} + {b^2}} \over b}} \right)$$
2
IIT-JEE 1999
MCQ (Single Correct Answer)
+2
-0.5
The curve described parametrically by $$x = {t^2} + t + 1,$$ $$y = {t^2} - t + 1 $$ represents
A
a pair of straight lines
B
an ellipse
C
a parabola
D
a hyperbola
3
IIT-JEE 1999
MCQ (Single Correct Answer)
+2
-0.5
If $$x$$ $$=$$ $$9$$ is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$ then the equation of the vcorresponding pair of tangents is
A
$$9{x^2} - 8{y^2} + 18x - 9 = 0$$
B
$$9{x^2} - 8{y^2} - 18x + 9 = 0$$
C
$$9{x^2} - 8{y^2} - 18x - 9 = 0$$
D
$$9{x^2} - 8{y^2} + 18x + 9 = 0$$
4
IIT-JEE 1999
MCQ (More than One Correct Answer)
+3
-0.75
On the ellipse $$4{x^2} + 9{y^2} = 1,$$ the points at which the tangents are parallel to the line $$8x = 9y$$ are
A
$$\left( {{2 \over 5},{1 \over 5}} \right)$$
B
$$\left( -{{2 \over 5},{1 \over 5}} \right)$$
C
$$\left( -{{2 \over 5},-{1 \over 5}} \right)$$
D
$$\left( {{2 \over 5},-{1 \over 5}} \right)$$
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