1
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

A point moves so that the sum of its distances from $$(a e, 0)$$ and $$(-a e, 0)$$ is $$2 a$$, then the equation to its locus, where $$b^2=a^2\left(1-e^2\right)$$ is

A
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
B
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
C
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$$
D
$$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$$
2
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

The point to which the origin should be shifted in order to eliminate the $$x$$ and $$y$$ terms from the equation $$9 x^2+4 y^2+10 x+12 y+1=0$$ is

A
$$\left(\frac{5}{9}, \frac{3}{2}\right)$$
B
$$\left(\frac{-5}{2}, \frac{-3}{9}\right)$$
C
$$\left(\frac{-5}{9}, \frac{-3}{2}\right)$$
D
$$\left(\frac{-3}{2}, \frac{-5}{9}\right)$$
3
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $$A(1,3)$$ and $$C(7,5)$$ are two opposite vertices of a square, then find the equation of a side passing through $$A$$.

A
$$x=y$$
B
$$x-2 y+1=0$$
C
$$x-3 y+8=0$$
D
$$2 x-y+1=0$$
4
AP EAPCET 2021 - 20th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$C$$ is the centroid of the triangle with vertices $$(3,-1),(1,3)$$ and $$(2,4)$$. Let $$P$$ be the point of intersection of the lines $$x+3 y-1=0$$ and $$3 x-y+1=0$$. Then a line which passes through both points $$C$$ and $$P$$ would also passes through the point .......

A
$$(-9,-7)$$
B
$$(-9,-6)$$
C
$$(7,6)$$
D
$$(9,7)$$