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 · 25 Jan · Shift 2 · Q22

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 · 25 Jan · Shift 2 · Q22

Parabola question

2023 · 25 Jan · Shift 2 · Q22

JEE MainMathematicsParabolaMCQ+4 / −1

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=6x

. The locus of its circumcentre is :

A
$4{y^2} - 18y - 3x - 18 = 0$
B
$4{y^2} + 18y + 3x + 18 = 0$
C
$4{y^2} - 18y + 3x + 18 = 0$
D
$4{y^2} - 18y - 3x + 18 = 0$

View written solutionFree

Correct answer: C

Set up the variable triangle

The two fixed sides are: $x=0 \quad \text{and} \quad y=3$ These intersect at $A=(0,3).$

The third side is a variable tangent to the parabola $y^2=6x.$

For the parabola $y^2=4ax$ , we have $4a=6 \Rightarrow a=\frac{3}{2}$ . A tangent in slope form is: $y=mx+\frac{a}{m}=mx+\frac{3}{2m}.$

So let the tangent be $y=mx+\frac{3}{2m}.$

Find the other two vertices of the triangle

The tangent meets $x=0$ at: $B=\left(0,\frac{3}{2m}\right).$

It meets $y=3$ at: $3=mx+\frac{3}{2m}$ $mx=3-\frac{3}{2m}$ $x=\frac{3}{m}-\frac{3}{2m^2}.$ Hence, $C=\left(\frac{3}{m}-\frac{3}{2m^2},\,3\right).$

So triangle $ABC$ has:

$A=(0,3)$
$B=\left(0,\frac{3}{2m}\right)$
$C=\left(\frac{3}{m}-\frac{3}{2m^2},3\right)$

Observe that the triangle is right-angled at $A$

$AB$ lies on $x=0$ , a vertical line.
$AC$ lies on $y=3$ , a horizontal line.

Therefore, $\angle A=90^\circ.$

For a right triangle, the circumcentre is the midpoint of the hypotenuse $BC$ .

Let the circumcentre be $P(h,k)$ . Then $P=\text{midpoint of }B\text{ and }C.$

So,

=\frac{3}{2m}-\frac{3}{4m^2},$$ $$k=\frac{1}{2}\left(\frac{3}{2m}+3\right) =\frac{3}{4m}+\frac{3}{2}. $$ --- 4. **Eliminate the parameter $m$** Let the circumcentre coordinates be $(x,y)$. Then $$x=\frac{3}{2m}-\frac{3}{4m^2},$$ $$y=\frac{3}{4m}+\frac{3}{2}. $$ From the second equation, $$y-\frac{3}{2}=\frac{3}{4m}$$ $$\frac{1}{m}=\frac{4}{3}\left(y-\frac{3}{2}\right)=\frac{4y-6}{3}. $$ Then $$\frac{1}{m^2}=\left(\frac{4y-6}{3}\right)^2. $$ Now substitute into the expression for $x$: $$x=\frac{3}{2}\cdot \frac{1}{m}-\frac{3}{4}\cdot \frac{1}{m^2}.$$ Using $\frac{1}{m}=\frac{4y-6}{3}$, $$x=\frac{3}{2}\cdot \frac{4y-6}{3}-\frac{3}{4}\left(\frac{4y-6}{3}\right)^2.$$ Simplify: $$x=\frac{4y-6}{2}-\frac{1}{12}(4y-6)^2$$ $$x=2y-3-\frac{1}{12}(16y^2-48y+36).$$ So, $$x=2y-3-\left(\frac{4}{3}y^2-4y+3\right)$$ $$x=-\frac{4}{3}y^2+6y-6.$$ Multiply by $3$: $$3x=-4y^2+18y-18$$ Rearrange: $$4y^2-18y+3x+18=0.$$ --- 5. **Match with the options** This is exactly: $$\boxed{4y^2-18y+3x+18=0}$$ which is **Option C**. --- 6. **Comparison with stored correct answer** Stored correct answer: **C** Derived answer: **C** So they agree.

More from Parabola

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
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $3 x + y - 29 = 0$ , is $x^{2} + a y^{2} + b x y + c x + d y + k = 0$ , then $a + b + c + d + k$ is equal to :2022 · MCQ
If vertex of a parabola is (2, $-$ 1) and the equation of its directrix is 4x $-$ 3y = 21, then the length of its latus rectum is :2022 · MCQ
Let the focal chord of the parabola $P : y^{2} = 4 x$ along the line $L : y = m x + c, m > 0$ meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $H : x^{2} - y^{2} = 4$ …2022 · MCQ