New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Parabola question

2023 · 1 Feb · Shift 2 · Q40

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Parabola
2023 · 1 Feb · Shift 2 · Q40

Parabola question

2023 · 1 Feb · Shift 2 · Q40

JEE MainMathematicsParabolaNumerical+4 / −1

If the

x

-intercept of a focal chord of the parabola

y^{2}=8x+4y+4

is 3, then the length of this chord is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 16

Write the parabola in standard form

Given: $y^2=8x+4y+4$

Rearrange and complete the square in $y$ : $y^2-4y=8x+4$ $y^2-4y+4=8x+8$ $(y-2)^2=8(x+1)$

This is of the form $(y-k)^2=4a(x-h)$ with

\quad k=2, \quad 4a=8 \Rightarrow a=2.$$ So the **focus** is $$(h+a,k)=(1,2).$$ --- 2. **Equation of a focal chord whose x-intercept is 3** A focal chord passes through the focus $(1,2)$. Its $x$-intercept is $3$, so it passes through $(3,0)$. Hence the chord is the line through $(1,2)$ and $(3,0)$. Slope: $$m=\frac{0-2}{3-1}=-1$$ So the equation is $$y-2=-1(x-1)$$ $$x+y-3=0$$ or $$y=3-x.$$ --- 3. **Find the points where this line meets the parabola** Substitute $y=3-x$ into $$(y-2)^2=8(x+1).$$ Since $$y-2=(3-x)-2=1-x,$$ we get $$(1-x)^2=8(x+1).$$ Expand: $$x^2-2x+1=8x+8$$ $$x^2-10x-7=0.$$ So the two intersection points have $x$-coordinates satisfying $$x_1+x_2=10, \quad x_1x_2=-7.$$ Because $y=3-x$, their $y$-coordinates are $y_1=3-x_1$, $y_2=3-x_2$. --- 4. **Length of the chord** Along the line $y=3-x$, the slope is $-1$, so if the difference in $x$-coordinates is $|x_1-x_2|$, then the distance is $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$ But $$y_1-y_2=-(x_1-x_2),$$ so $$\text{length} = \sqrt{(x_1-x_2)^2+(x_1-x_2)^2} =\sqrt{2}\,|x_1-x_2|.$$ Now $$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=10^2-4(-7)=100+28=128.$$ Thus $$|x_1-x_2|=\sqrt{128}=8\sqrt{2}.$$ Therefore, $$\text{length}=\sqrt{2}\cdot 8\sqrt{2}=16.$$ --- 5. **Final Answer** The length of the focal chord is $$\boxed{16}.$$

More from Parabola

Let $R$ be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points $P$ and $Q$ . Let the point $G (10, 10)$ be the centroid of the triangle $P QR$ . If $c - m = 6$ , then $(P Q)^{2}$ is :2023 · MCQ
Let $PQ$ be a focal chord of the parabola $y^{2} = 36 x$ of length 100 , making an acute angle with the positive $x$ -axis. Let the ordinate of $P$ be positive and $M$ be the point on the line segment PQ such that…2023 · MCQ
The equations of two sides of a variable triangle are $x = 0$ and $y = 3$ , and its third side is a tangent to the parabola $y^{2} = 6 x$ . The locus of its circumcentre is :2023 · MCQ
The parabolas : $a x^{2} + 2 b x + cy = 0$ and $d x^{2} + 2 e x + f y = 0$ intersect on the line $y = 1$ . If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then :2023 · MCQ
A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :2022 · MCQ
Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = .2022 · Numerical
Let $x = 2 t$ , $y = \frac{t ^{2}}{3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $S A ⊥ B A$ , where A is any point on the conic. If k is the ordinate of the centroid of the $Δ$ SAB, then $t \to 1 lim k$ …2022 · MCQ
Let $P$ and $Q$ be any points on the curves $(x - 1)^{2} + (y + 1)^{2} = 1$ and $y = x^{2}$ , respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval :2022 · MCQ