Let $P$ be the point $(1, 0)$ and $Q$ be a point on the locus $y^2 = 8x$. The locus of the midpoint of $PQ$ is
- A$x^2 + 4y + 2 = 0$
- B$x^2 - 4y + 2 = 0$
- C$y^2 - 4x + 2 = 0$
- D$y^2 + 4x + 2 = 0$
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Similar Questions
The equation of the normals at the ends of the latus rectum of the parabola $y^{2} = 4ax$ is:
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Medium
View SolutionIf the normal to the parabola $y^2=4x$ at $P(1,2)$ meets the parabola again at $Q$,then the coordinates of $Q$ are
The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$
Consider the parabola $25[(x-2)^2+(y+5)^2]=(3x+4y-1)^2$. Match the characteristics of this parabola given in List-$I$ with their corresponding items in List-$II$.
List-$I$ List-$II$ $I$. Vertex $A$. $8$ $II$. Length of latus rectum $B$. $(\frac{29}{10}, \frac{-38}{10})$ $III$. Directrix $C$. $3x+4y-1=0$ $IV$. One end of the latus rectum $D$. $(\frac{-2}{5}, \frac{-16}{5})$ $E$. $6$
| List-$I$ | List-$II$ |
| $I$. Vertex | $A$. $8$ |
| $II$. Length of latus rectum | $B$. $(\frac{29}{10}, \frac{-38}{10})$ |
| $III$. Directrix | $C$. $3x+4y-1=0$ |
| $IV$. One end of the latus rectum | $D$. $(\frac{-2}{5}, \frac{-16}{5})$ |
| $E$. $6$ |
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