# ISI2014-DCG-27

69 views

Let $y^2-4ax+4a=0$ and $x^2+y^2-2(1+a)x+1+2a-3a^2=0$ be two curves. State which one of the following statements is true.

1. These two curves intersect at two points
2. These two curves are tangent to each other
3. These two curves intersect orthogonally at one point
4. These two curves do not intersect
in Geometry
retagged
2

@ankitgupta.1729

What could be the right way to solve this?

I am trying by putting $y^2$ value from equation 1 to equation 2.

1

@`JEET

yeah, it is right way or subtract one equation from other to eliminate 'y'. Here, answer should be (A).

1
Thanks.

## Related questions

1
48 views
The area under the curve $x^2+3x-4$ in the positive quadrant and bounded by the line $x=5$ is equal to $59 \frac{1}{6}$ $61 \frac{1}{3}$ $40 \frac{2}{3}$ $72$
Let the following two equations represent two curves $A$ and $B$. $A: 16x^2+9y^2=144\:\: \text{and}\:\: B:x^2+y^2-10x=-21$ Further, let $L$ and $M$ be the tangents to these curves $A$ and $B$, respectively, at the point $(3,0)$. Then the angle between these two tangents, $L$ and $M$, is $0^{\circ}$ $30^{\circ}$ $45^{\circ}$ $90^{\circ}$
If $A(t)$ is the area of the region bounded by the curve $y=e^{-\mid x \mid}$ and the portion of the $x$-axis between $-t$ and $t$, then $\underset{t \to \infty}{\lim} A(t)$ equals $0$ $1$ $2$ $4$
The shaded region in the following diagram represents the relation $y\:\leq\: x$ $\mid \:y\mid \:\leq\: \mid x\:\mid$ $y\:\leq\: \mid x\:\mid$ $\mid \:y\mid\: \leq\: x$