a) Prove that the locus of the middle points of the chords of parabola y^2=4ax which passes through the focus is a parabola. [4] OR Find the eccentricity, the coordinate of the centre and the foci…

a) Prove that the locus of the middle points of the chords of parabola $y^2=4ax$ which passes through the focus is a parabola.

Step 1: Understand the given parabola and its focus

The given parabola is: $$y^2=4ax$$ This is a standard right-opening parabola with:

• Vertex at $(0,0)$
• Focus at $(a,0)$

Step 2: Let the chord pass through the focus

Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the endpoints of a chord passing through the focus $(a,0)$. Since $A$ and $B$ lie on the parabola, they satisfy: $$y_1^2=4a{x}_1\text{and}{y}_2^2=4a{x}_2$$

Since the chord passes through $(a,0)$, the condition for collinearity is: $$\frac{y_1-0}{x_1-a}=\frac{y_2-0}{x_2-a}$$ Simplifying: $$y_1(x_2-a)=y_2(x_1-a)$$ $$y_1{x}_2-a{y}_1=y_2{x}_1-a{y}_2$$ $$y_1{x}_2-y_2{x}_1=a(y_1-y_2)\text{(Equation 1)}$$

Step 3: Find the midpoint of the chord

Let $(h,k)$ be the midpoint of chord $AB$. Then: $$h=\frac{x_1+x_2}{2},k=\frac{y_1+y_2}{2}$$ Thus: $$x_1+x_2=2h,y_1+y_2=2k$$

Step 4: Use the parabola condition

From $y_1^2=4a{x}_1$ and $y_2^2=4a{x}_2$, subtract: $$y_1^2-y_2^2=4a(x_1-x_2)$$ $$(y_1-y_2)(y_1+y_2)=4a(x_1-x_2)$$ Substitute $y_1+y_2=2k$: $$(y_1-y_2)(2k)=4a(x_1-x_2)$$ $$(y_1-y_2)=\frac{2a(x_1-x_2)}{k}\text{(Equation 2)}$$

Step 5: Substitute into Equation 1

From Equation 1: $$y_1{x}_2-y_2{x}_1=a(y_1-y_2)$$ Substitute $y_1-y_2$ from Equation 2: $$y_1{x}_2-y_2{x}_1=a\left(\frac{2a(x_1-x_2)}{k}\right)$$ $$y_1{x}_2-y_2{x}_1=\frac{2a^2(x_1-x_2)}{k}$$ Rearrange: $$k(y_1{x}_2-y_2{x}_1)=2a^2(x_1-x_2)$$

Now, express $y_1$ and $y_2$ in terms of $x_1$ and $x_2$: From $y_1^2=4a{x}_1$ and $y_2^2=4a{x}_2$, we can write: $$y_1=2\sqrt{a{x}_1},y_2=2\sqrt{a{x}_2}$$ But this complicates the algebra. Instead, use parametric coordinates for the parabola.

Alternative Approach: Parametric Coordinates

Let the endpoints of the chord be: $$A=(a{t}_1^2,2a{t}_1),B=(a{t}_2^2,2a{t}_2)$$ Since the chord passes through the focus $(a,0)$, the condition for collinearity is: $$\frac{2a{t}_1-0}{a{t}_1^2-a}=\frac{2a{t}_2-0}{a{t}_2^2-a}$$ Simplify: $$\frac{2t_1}{t_1^2-1}=\frac{2t_2}{t_2^2-1}$$ $$t_1(t_2^2-1)=t_2(t_1^2-1)$$ $$t_1{t}_2^2-t_1=t_2{t}_1^2-t_2$$ $$t_1{t}_2^2-t_2{t}_1^2=t_1-t_2$$ $$t_1{t}_2(t_2-t_1)=-(t_2-t_1)$$ Assuming $t_1\ne t_2$, divide both sides by $(t_2-t_1)$: $$t_1{t}_2=-1$$

Step 6: Find the midpoint coordinates

The midpoint $(h,k)$ is: $$h=\frac{a{t}_1^2+a{t}_2^2}{2}=\frac{a(t_1^2+t_2^2)}{2}$$ $$k=\frac{2a{t}_1+2a{t}_2}{2}=a(t_1+t_2)$$

From $t_1{t}_2=-1$, we can express $t_1^2+t_2^2$ as: $$t_1^2+t_2^2=(t_1+t_2{)}^2-2t_1{t}_2=(t_1+t_2{)}^2+2$$ Let $t_1+t_2=m$. Then: $$h=\frac{a(m^2+2)}{2},k=am$$ From $k=am$, we get $m=\frac{k}{a}$. Substitute into $h$: $$h=\frac{a\left({\left(\frac{k}{a}\right)}^2+2\right)}{2}=\frac{a\left(\frac{k^2}{a^2}+2\right)}{2}=\frac{k^2}{2a}+a$$ Rearrange: $$h=\frac{k^2}{2a}+a$$ $$k^2=2a(h-a)$$ This is the equation of a parabola in terms of $h$ and $k$. Replacing $h$ with $x$ and $k$ with $y$, we get: $$y^2=2a(x-a)$$ This is the equation of a parabola with vertex at $(a,0)$ and axis parallel to the x-axis.

Conclusion: The locus of the midpoints is indeed a parabola.

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OR

Find the eccentricity, the coordinate of the centre and the foci of the curve $x^2+4y^2-4x+24y+24=0$.

Step 1: Rewrite the equation in standard form

The given equation is: $$x^2+4y^2-4x+24y+24=0$$ Group $x$ and $y$ terms: $$x^2-4x+4y^2+24y=-24$$ Complete the square for $x$ and $y$:

For $x$: $$x^2-4x=(x^2-4x+4)-4=(x-2{)}^2-4$$

For $y$: $$4y^2+24y=4(y^2+6y)=4(y^2+6y+9-9)=4[(y+3{)}^2-9]=4(y+3{)}^2-36$$

Substitute back: $$(x-2{)}^2-4+4(y+3{)}^2-36=-24$$ $$(x-2{)}^2+4(y+3{)}^2-40=-24$$ $$(x-2{)}^2+4(y+3{)}^2=16$$ Divide by 16: $$\frac{(x-2{)}^2}{16}+\frac{(y+3{)}^2}{4}=1$$

Step 2: Identify the conic section

The equation is of the form: $$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$$ where $a^2=16$ and $b^2=4$. Since $a>b$, this is an ellipse.

Step 3: Find the centre, eccentricity, and foci

• Centre: $(h,k)=(2,-3)$
• Semi-major axis ($a$): $a=4$
• Semi-minor axis ($b$): $b=2$
• Eccentricity ($e$): $$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{16}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$$
• Foci: The distance of each focus from the centre is $ae$: $$ae=4\times \frac{\sqrt{3}}{2}=2\sqrt{3}$$ Thus, the foci are at: $$(2\pm 2\sqrt{3},-3)$$

Final Results:

• Centre: $(2,-3)$
• Eccentricity: $\frac{\sqrt{3}}{2}$
• Foci: $(2+2\sqrt{3},-3)$ and $(2-2\sqrt{3},-3)$

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b) Find the equation of the plane passing through the midpoint of $P(-2,5,1)$ and $Q(6,1,5)$ and perpendicular to the line joining them.

Step 1: Find the midpoint of $P$ and $Q$

$$\text{Midpoint}=\left(\frac{-2+6}{2},\frac{5+1}{2},\frac{1+5}{2}\right)=(2,3,3)$$

Step 2: Find the direction vector of $PQ$

$$\overrightarrow{PQ}=Q-P=(6-(-2),1-5,5-1)=(8,-4,4)$$

Step 3: Use the point-normal form of the plane equation

The plane passes through $(2,3,3)$ and has normal vector $\vec{n}=(8,-4,4)$. The equation is: $$8(x-2)-4(y-3)+4(z-3)=0$$ Simplify: $$8x-16-4y+12+4z-12=0$$ $$8x-4y+4z-16=0$$ Divide by 4: $$2x-y+z-4=0$$

Final Equation of the Plane: $$\boxed{2x-y+z-4=0}$$