a) Find condition that a line y = mx + c is tangent to the parabola y^2 = 4ax . [4] OR Find the vertices and foci of the curve (x+6)^24 + y^236 = 1 . b) Find the equation of the plane through the…

a) Find the condition that a line $y=mx+c$ is tangent to the parabola $y^2=4ax$.

Step 1: Understand the problem

We need to find the condition under which the line $y=mx+c$ touches the parabola $y^2=4ax$ at exactly one point (i.e., the line is tangent to the parabola).

Step 2: Substitute the line equation into the parabola equation

The line is given by: $$y=mx+c$$

The parabola is given by: $$y^2=4ax$$

Substitute $y$ from the line into the parabola: $$(mx+c{)}^2=4ax$$ $$m^2{x}^2+2mcx+c^2=4ax$$ $$m^2{x}^2+(2mc-4a)x+c^2=0$$

Step 3: Condition for tangency

For the line to be tangent to the parabola, the quadratic equation in $x$ must have exactly one solution. This means the discriminant must be zero.

The general form of a quadratic equation is: $$A{x}^2+Bx+C=0$$

Here:

• $A=m^2$
• $B=2mc-4a$
• $C=c^2$

The discriminant $D$ is: $$D=B^2-4AC$$

Set $D=0$ for tangency: $$(2mc-4a{)}^2-4\cdot m^2\cdot c^2=0$$

Step 4: Simplify the equation

Expand $(2mc-4a{)}^2$: $$4m^2{c}^2-16amc+16a^2-4m^2{c}^2=0$$ $$-16amc+16a^2=0$$ $$16a(-mc+a)=0$$

Since $a\ne 0$ (as $y^2=4ax$ is a valid parabola), we have: $$-mc+a=0$$ $$mc=a$$

Final condition for tangency

The condition for the line $y=mx+c$ to be tangent to the parabola $y^2=4ax$ is: $$\boxed{c=\frac{a}{m}}$$

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OR

Find the vertices and foci of the curve $\frac{(x+6{)}^2}{4}+\frac{y^2}{36}=1$.

Step 1: Identify the type of curve

The given equation is: $$\frac{(x+6{)}^2}{4}+\frac{y^2}{36}=1$$

This is the standard form of an ellipse: $$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$$ where:

• $(h,k)$ is the center,
• $a$ is the semi-major or semi-minor axis,
• $b$ is the other axis.

Step 2: Rewrite in standard form

Compare with the given equation: $$\frac{(x-(-6){)}^2}{2^2}+\frac{y^2}{6^2}=1$$

Here:

• Center $(h,k)=(-6,0)$,
• $a=2$ (since $a^2=4$),
• $b=6$ (since $b^2=36$).

Since $b>a$, the major axis is along the y-axis.

Step 3: Find the vertices

For an ellipse with major axis along the y-axis:

• Vertices are at $(h,k\pm b)$.

Thus: $$\text{Vertices}=(-6,0\pm 6)=(-6,6)\text{ and }(-6,-6)$$

Step 4: Find the foci

The distance of each focus from the center is given by: $$c=\sqrt{b^2-a^2}=\sqrt{36-4}=\sqrt{32}=4\sqrt{2}$$

Since the major axis is along the y-axis, the foci are at: $$(h,k\pm c)=(-6,0\pm 4\sqrt{2})$$

Thus: $$\text{Foci}=(-6,4\sqrt{2})\text{ and }(-6,-4\sqrt{2})$$

Final answer

• Vertices: $\boxed{(-6,6)}$ and $\boxed{(-6,-6)}$
• Foci: $\boxed{(-6,4\sqrt{2})}$ and $\boxed{(-6,-4\sqrt{2})}$

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b) Find the equation of the plane through the point $(2,-3,1)$ and perpendicular to the line joining the two points $(3,4,-1)$ and $(2,-1,5)$.

Step 1: Find the direction vector of the line

The line joins the points $A(3,4,-1)$ and $B(2,-1,5)$.

The direction vector $\overrightarrow{AB}$ is: $$\overrightarrow{AB}=B-A=(2-3,-1-4,5-(-1))=(-1,-5,6)$$

Step 2: Equation of the plane

The plane is perpendicular to the line $AB$, so the normal vector of the plane is the same as the direction vector of the line: $$\vec{n}=(-1,-5,6)$$

The general equation of a plane with normal vector $\vec{n}=(A,B,C)$ passing through $(x_0,y_0,z_0)$ is: $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$

Substitute $\vec{n}=(-1,-5,6)$ and the point $(2,-3,1)$: $$-1(x-2)-5(y+3)+6(z-1)=0$$

Step 3: Simplify the equation

$$-(x-2)-5(y+3)+6(z-1)=0$$ $$-x+2-5y-15+6z-6=0$$ $$-x-5y+6z-19=0$$ $$x+5y-6z+19=0$$

Final equation of the plane

$$\boxed{x+5y-6z+19=0}$$