1
JEE Advanced 2024 Paper 1 Online
+3
-1

Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. Two tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac{3}{2}$, then which of the following options is correct?

A
$q=2, p=3 \sqrt{3}$
B
$q=2, p=4 \sqrt{3}$
C
$q=1, p=5 \sqrt{3}$
D
$q=1, p=6 \sqrt{3}$
2
JEE Advanced 2022 Paper 1 Online
+3
-1

Consider the ellipse

$$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$$$Let$H(\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through$H$parallel to the$y$-axis crosses the ellipse and its auxiliary circle at points$E$and$F$respectively, in the first quadrant. The tangent to the ellipse at the point$E$intersects the positive$x$-axis at a point$G$. Suppose the straight line joining$F$and the origin makes an angle$\phi$with the positive$x$-axis. List-I List-II (I) If$\phi=\frac{\pi}{4}$, then the area of the triangle$F G H$is (P)$\frac{(\sqrt{3}-1)^{4}}{8}$(II) If$\phi=\frac{\pi}{3}$, then the area of the triangle$F G H$is (Q) 1 (III) If$\phi=\frac{\pi}{6}$, then the area of the triangle$F G H$is (R)$\frac{3}{4}$(IV) If$\phi=\frac{\pi}{12}$, then the area of the triangle$F G H$is (S)$\frac{1}{2 \sqrt{3}}$(T)$\frac{3 \sqrt{3}}{2}$The correct option is: A$(\mathrm{I}) \rightarrow(\mathrm{R}) ;(\mathrm{II}) \rightarrow(\mathrm{S}) ;(\mathrm{III}) \rightarrow(\mathrm{Q}) ;(\mathrm{IV}) \rightarrow(\mathrm{P})$B (I)$\rightarrow$(R); (II)$\rightarrow(\mathrm{T}) ;(\mathrm{III}) \rightarrow(\mathrm{S}) ;(\mathrm{IV}) \rightarrow(\mathrm{P})$C (I)$\rightarrow(\mathrm{Q}) ;(\mathrm{II}) \rightarrow(\mathrm{T}) ;(\mathrm{III}) \rightarrow(\mathrm{S}) ;(\mathrm{IV}) \rightarrow(\mathrm{P})$D (I)$\rightarrow$(Q); (II)$\rightarrow$(S); (III)$\rightarrow$(Q); (IV)$\rightarrow\$ (P)
3
JEE Advanced 2018 Paper 1 Offline
+3
-1
Let S be the circle in the XY-plane defined the equation x2 + y2 = 4.

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve
A
(x + y)2 = 3xy
B
x2/3 + y2/3 = 24/3
C
x2 + y2 = 2xy
D
x2 + y2 = x2y2
4
JEE Advanced 2016 Paper 2 Offline
+3
-0
Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

If the tangents to the ellipse at $$M$$ and $$N$$ meet at $$R$$ and the normal to the parabola at $$M$$ meets the $$x$$-axis at $$Q$$, then the ratio of area of the triangle $$MQR$$ to area of the quadrilateral $$M{F_1}N{F_2}$$is

A
$$3:4$$
B
$$4:5$$
C
$$5:8$$
D
$$2:3$$
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