The point of intersection of the lines (3)kx+ky−43=0 and 3x−y−4(3)k=0.
k=3x+y43=433x−y
⇒3x2−y2=48
⇒16x2−48y2=1
Which is a hyperbola.
b2=a2(e2−1)
Now, 48=16(e2−1)
⇒e=4=2
The locus of the point of intersection of the lines (3)kx+ky−43=0 and 3x−y−4(3)k=0 is a conic, whose eccentricity is
Held on 25 Feb 2021 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the image of parabola $x^{2}=4 y$, in the line $x-y=1$ be $(y+a)^{2}=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to
The distance between the points (3, 4) and (6, 8) is:
If the chord joining the points $\mathrm{P}_{1}\left(x_{1}, y_{1}\right)$ and $\mathrm{P}_{2}\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2}-y_{1} y_{2}$ is equal to
The distance between the parallel lines 3x + 4y - 7 = 0 and 3x + 4y + 8 = 0 is:
Let a point $A$ lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :
Work through every JEE Main Coordinate Geometry PYQ, year by year.