JEE MAIN Parabola Previous Year Questions (PYQs) – Page 1 of 5
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If the line 3x - 2y + 12 = 0 intersects the parabola 4y = 3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to:
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Solution
Given line: $3x-2y+12=0 \Rightarrow y=\frac{3x+12}{2}$
Parabola: $4y=3x^2 \Rightarrow y=\frac{3x^2}{4}$
Equate:
$\frac{3x^2}{4}=\frac{3x+12}{2}$
$\Rightarrow 3x^2=6x+24$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$
Points: $A(4,12),\ B(-2,3)$
Vertex $V(0,0)$
Slope $VA=\frac{12}{4}=3,\quad VB=\frac{3}{-2}=-\frac{3}{2}$
$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{3+\frac{3}{2}}{1- \frac{9}{2}}\right| =\left|\frac{\frac{9}{2}}{-\frac{7}{2}}\right| =\frac{9}{7}$
$\theta=\tan^{-1}\left(\frac{9}{7}\right)$
$\boxed{\tan^{-1}\left(\frac{9}{7}\right)}$
Parabola: $4y=3x^2 \Rightarrow y=\frac{3x^2}{4}$
Equate:
$\frac{3x^2}{4}=\frac{3x+12}{2}$
$\Rightarrow 3x^2=6x+24$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$
Points: $A(4,12),\ B(-2,3)$
Vertex $V(0,0)$
Slope $VA=\frac{12}{4}=3,\quad VB=\frac{3}{-2}=-\frac{3}{2}$
$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{3+\frac{3}{2}}{1- \frac{9}{2}}\right| =\left|\frac{\frac{9}{2}}{-\frac{7}{2}}\right| =\frac{9}{7}$
$\theta=\tan^{-1}\left(\frac{9}{7}\right)$
$\boxed{\tan^{-1}\left(\frac{9}{7}\right)}$
The area (in sq. units) of the region described by $\{(x,y):y^{2}\le2x,\;y\ge4x-1\}$ is:
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Solution
Let $PQ$ be a chord of the parabola $y^{2}=12x$ and the midpoint of $PQ$ be at $(4,1)$.
Then, which of the following points lies on the line passing through the points $P$ and $Q$?
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Solution
The axis of a parabola is the line $y=x$ and its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ units from the origin, respectively. If the point $(1,k)$ lies on the parabola, then a possible value of $k$ is:
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Solution
The length of the chord of the parabola $x^2=4y$ having equation $x-\sqrt{2}\,y+4\sqrt{2}=0$ is –
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Solution
Let the shortest distance from $(a,0)$, $a>0$, to the parabola $y^{2}=4x$ be $4$.
Then the equation of the circle passing through the point $(a,0)$ and the focus of the parabola, with centre on the axis of the parabola, is:
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Solution
Let P be the parabola, whose focus is (-2,1) and directrix is 2x+y+2=0. Then the sum of the ordinates of the points on P, whose abscissa is -2, is
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Solution
If the chord joining the points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ on the parabola $y^2 = 12x$ subtends a right angle at the vertex of the parabola, then $x_1x_2 - y_1y_2$ is equal to
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Solution
$(x_1,y_1) = (3t_1^2, 6t_1),; (x_2,y_2) = (3t_2^2, 6t_2)$
$t_1t_2 = -4$
$x_1x_2 = 9(t_1t_2)^2,; y_1y_2 = 36t_1t_2$
$x_1x_2 - y_1y_2 = 9(16) - 36(-4)$
$= 144 + 144 = 288$
$f:R \to R$ be defined as$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.$ Let A = {x $ \in $ R : f is increasing}. Then A is equal to :
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Solution
If the shortest distance of the parabola $y^2=4x$ from the centre of the circle
$x^2+y^2-4x-16y+64=0$ is $d$, then $d^2$ is equal to:
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Solution
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