
Point of intersection of lines
x−y=1 and 2x+y=3
O is (34,31)
Slope of OP=34−131+1=3134=4
Slope of tangent =−41
⇒ Equation of tangent :y+1=−41(x−1)
⇒4y+4=−x+1
⇒x+4y+3=0
JEE Main 2016 — Mathematics Coordinate Geometry
Equation of the tangent to the circle, at the point (1,−1), whose center, is the point of intersection of the straight lines x−y=1 and 2x+y=3 is:
Held on 10 Apr 2016 · Verified 6 Jul 2026.
x+4y+3=0
3x−y−4=0
x−3y−4=0
4x+y−3=0
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
Let the domain of the function $f(x)=\log _{3} \log _{5} \log _{7}\left(9 x-x^{2}-13\right)$ be the interval $(\mathrm{m}, \mathrm{n})$. Let the hyperbola $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 \mathrm{~m}}{3}$. Then $\mathrm{b}^{2}-\mathrm{a}^{2}$ is equal to :
The distance between the points (3, 4) and (6, 8) is:
If P is a point on the circle $x^{2}+y^{2}=4, \mathrm{Q}$ is a point on the straight line $5 x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P 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.