Given equation of lines are (a3+3)x+ay+a−3=0 and (a5+2)x+(a+2)y+2a+3=0 (a real) Since point of intersection of lines lies on y-axis. ∴ Put x=0 in each equation, we get ay+a−3=0 and (a+2)y+2a+3=0 On solving these we get (a+2)(a−3)−a(2a+3)=0⇒a2−a−6−2a2−3a=0⇒−a2−4a−6=0⇒a2+4a+6=0⇒a=2−4±16−24=2−4±−8( not real) This shows that the point of intersection of the lines lies on the y-axis for no value of ' a '.
The point of intersection of the lines (a^3+3 ) x+a y+a-3=0 and (a^5+2 ) x+(a+2) y+2 a+3=0 (a real) lies on the y-axis for
JEE Main 2012 — Mathematics Coordinate Geometry
The point of intersection of the lines (a3+3)x+ay+a−3=0 and (a5+2)x+(a+2)y+2a+3=0 (a real) lies on the y-axis for
Held on 7 May 2012 · Verified 6 Jul 2026.
Options
- A
no value of a
- B
more than two values of a
- C
exactly one value of a
- D
exactly two values of a
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Coordinate Geometry in other years
More JEE Main Coordinate Geometry questions
Let $A,B$ be points on the two half-lines $x-\sqrt{3}|y|=\alpha$, $\alpha>0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ=\dfrac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\dfrac{\alpha^2}{R}$ is equal to ______
If a straight line drawn through the point of intersection of the lines $4x+3y-1=0$ and $3x+4y-1=0$, meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:
Let $C$ be a circle having centre in the first quadrant and touching the $x$-axis at a distance of $3$ units from the origin. If the circle $C$ has an intercept of length $6\sqrt{3}$ on $y$-axis, then the length of the chord of the circle $C$ on the line $x - y = 3$ is :
Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$. Then the area of the triangle OPQ equals:
Let the vertex $A$ of a triangle $ABC$ be $(1, 2)$, and the mid-point of the side $AB$ be $(5, -1)$. If the centroid of this triangle is $(3, 4)$ and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:
Other Mathematics topics for JEE Main
Keep practicing Coordinate Geometry
Work through every JEE Main Coordinate Geometry PYQ, year by year.