Conic Sections Test - 51

$${\textbf{Step 1: Compare the coordinates of given points with the end points of line.}}$$

                $${\text{Given end point of diameter are}}$$$$\left( { - 6,3} \right)$$$${\text{and}}$$$$\left( {6,4} \right)$$

                $${\text{Comparing with}}$$ $$\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$$

                $${\text{Here,}}$$$${x_1} =  - 6,{y_1} = 3,{x_2} = 6,{y_2} = 3$$

$${\textbf{Step 2: Put these values in eqn of mid-point of a line.}}$$

                $${\text{Now, Centre of a circle is mid - point of the diameter}}{\text{.}}$$

                $${\text{Mid-point of a line is given by=}}$$ $$\left( {\frac{{{x_1} + {x_2}}}{2}} \right),\left( {\frac{{{y_1} + {y_2}}}{2}} \right)$$

                $${\text{Therefore, centre of circle is   =}}$$$$\left( {\frac{{ - 6 + 6}}{2}} \right),\left( {\frac{{3 + 4}}{2}} \right)$$

                                                                      $$ = \left( {0,\frac{7}{2}} \right)$$

$${\textbf{Final Answer: Hence, the required answer is}}$$ $$\mathbf{\left( {0,\frac{7}{2}} \right).}$$

$$b=\dfrac{5}{2}$$
$$(2,1)$$ lies on $$\dfrac{x^{2}}{a^{2}}-\dfrac{4y^{2}}{25}=1$$
$$\therefore \dfrac{4}{a^{2}}=\dfrac{29}{25}$$

Equation of circle having centre of origin $$(0,0)$$ and radius $$=r$$,

$$S_{1};x^{2}+y^{2}=r^{2}----(1)$$

$$\therefore f(m_{1},\dfrac{1}{m_{2}}),f(m_{2},\dfrac{1}{m_{2}}),--f(m_{4},\dfrac{1}{m_{4}})$$
These points lie on $$S_{1}$$.

Let $$\displaystyle f(m, \frac{1}{m})$$ is point lie on S_{1},

$$m^{2}+\dfrac{1}{m^{2}}=r^{2}$$

$$m^{4}+1-r^{2}m^{2}=0$$

$$m^{4}-1-r^{2}m^{2}+1=0$$

$$m_{1},m_{2},m_{3}$$ and $$m_{4}$$ are roots of this equation

So, $$m_{1}m_{2}m_{3}m_{4} =1$$

$$136 (x^{2}+y^{2})=(5x+3y+7)^{2}$$

$$ \Rightarrow \displaystyle (x-0)^2+(y-0)^2= \dfrac{1}{136}(5x+3y+7)^{2}$$

$$ \Rightarrow \displaystyle (x-0)^2+(y-0)^2= \left( \frac{1}{2}\right)^2 \left(\frac{5x+3y+7}{\sqrt{34}} \right)^2$$

From the above form, we get focus of the conic has coordinates $$(0,0)$$, and directrix has equation $$5x+3y+7=0$$ and eccentricity $$e=\dfrac{1}{2}$$

Distance between focus and directrix $$=\dfrac{7}{\sqrt{34}}$$

Length of latus rectum$$=2\dfrac{b^2}{a}=2 \times e \times$$ distance bet. focus and directrix$$=2 \times \dfrac{1}{2} \times \dfrac{7}{\sqrt{34}}$$

$$= \dfrac{7}{\sqrt{34}}$$

Hence, option B.

$$x^2 + y^2 - 2xy - 8x - 8y + 32 = 0$$
$$\Rightarrow (x-y)^2 = 8 (x + y - 4)$$
is a parabola whose axis is $$x - y = 0$$ and the tangent at the vertex is $$x + y - 4 = 0$$.
Also, when $$y = 0$$, we have
$$x^2 - 8x + 32 = 0$$
which gives no real values of $$x$$.
when $$x = 0$$, we have $$y^2 - 8y + 32 = 0$$ which gives no real values of $$y$$.
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.

$$y^2=4a^{2}\sin^{2}{\theta}=$$
        $$=4a^{2}-4ax=-4a(x-a)$$
Now, since $$0\le\cos^{2}{\theta}\le{1}$$ 
$$\Rightarrow 0\le x \le a$$             
Also, $$-1\le\sin{\theta}\le{1}$$
$$\Rightarrow -2a\le y \le 2a$$
So, the locus of $$P$$ is a part of parabola.

Any point on the lines $$7x + y + 3 = 0$$ is $$Q (t, -3 -7 t),       t  \in R.$$
Now P $$(h, k)$$ is image of point Q in the line $$x - y +1 = 0$$
Then, $$\displaystyle \frac{h - t}{1} = \frac{k - (-3 - 7 t)}{-1}$$
         $$ = \displaystyle - \frac{2 (t - (-3 - 7t) + 1)}{1+1}$$
         $$ = -8t - 4$$
$$\Rightarrow       (h, k) \equiv (-7t - 4, t + 1)$$
This point lies on the circle $$x^2 + y^2 = 9$$
$$\Rightarrow (-7t - 4)^2 + (t + 1)^2 = 9$$
$$\Rightarrow 50 r^2 + 58t + 8 = 0$$
$$\Rightarrow 25r^2 + 19t + 4 = 0$$
$$ \Rightarrow (25 t + 4) (t + 1) = 0$$
$$\Rightarrow t = -4/25, t = 1$$
$$\Rightarrow (h, k) = \displaystyle \left ( -\frac{72}{25}, \frac{21}{25} \right ) $$ or $$(3, 0)$$