a) Find the condition that a line ax + by + c = 0 may be normal to the parabola y^2 = 4mx . [4] OR Find the vertices and foci of the ellipse (x+2)^216 + (y 5)^29 = 1 . b) Show that the plane 2x + 3y…

a) Find the condition that a line $ax+by+c=0$ may be normal to the parabola $y^2=4mx$.

Step 1: Understand the Parabola and Normal Line

The given parabola is $y^2=4mx$, which is a standard right-opening parabola with vertex at the origin $(0,0)$ and focus at $(m,0)$.

A normal to a curve at a point is a line perpendicular to the tangent at that point. To find the condition for the line $ax+by+c=0$ to be normal to the parabola, we need to:

1. Find the slope of the tangent to the parabola at any point $(x_1,y_1)$.
2. Determine the slope of the normal line (which is the negative reciprocal of the tangent slope).
3. Equate the slope of the given line to the slope of the normal and derive the condition.

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Step 2: Find the Slope of the Tangent to the Parabola

Differentiate $y^2=4mx$ implicitly with respect to $x$: $$2y\frac{dy}{dx}=4m⟹\frac{dy}{dx}=\frac{2m}{y}$$ At a point $(x_1,y_1)$ on the parabola, the slope of the tangent is: $$m_{\text{tangent}}=\frac{2m}{y_1}$$

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Step 3: Find the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent slope: $$m_{\text{normal}}=-\frac{y_1}{2m}$$

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Step 4: Express the Given Line in Slope-Intercept Form

The given line is $ax+by+c=0$. Rewrite it in slope-intercept form: $$by=-ax-c⟹y=-\frac{a}{b}x-\frac{c}{b}$$ The slope of this line is: $$m_{\text{line}}=-\frac{a}{b}$$

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Step 5: Equate the Slopes and Use the Parabola Condition

For the line to be normal to the parabola, its slope must equal the slope of the normal: $$-\frac{a}{b}=-\frac{y_1}{2m}⟹\frac{a}{b}=\frac{y_1}{2m}⟹y_1=\frac{2ma}{b}$$

Since $(x_1,y_1)$ lies on the parabola $y^2=4mx$, substitute $y_1$: $${\left(\frac{2ma}{b}\right)}^2=4m{x}_1⟹\frac{4m^2{a}^2}{b^2}=4m{x}_1⟹x_1=\frac{m{a}^2}{b^2}$$

Now, the point $(x_1,y_1)$ must satisfy the line equation $ax+by+c=0$: $$a\left(\frac{m{a}^2}{b^2}\right)+b\left(\frac{2ma}{b}\right)+c=0$$ Simplify: $$\frac{m{a}^3}{b^2}+\frac{2m{a}^2}{b}+c=0$$ Multiply through by $b^2$ to eliminate denominators: $$m{a}^3+2m{a}^{2}b+c{b}^2=0$$ Factor out $a$: $$a(m{a}^2+2mab+cb)=0$$ Since $a\ne 0$ (otherwise the line is not defined), we have: $$m{a}^2+2mab+cb=0$$ Divide by $m$ (assuming $m\ne 0$): $$a^2+2ab+\frac{cb}{m}=0$$ This is the required condition for the line $ax+by+c=0$ to be normal to the parabola $y^2=4mx$.

Final Condition: $$\boxed{a^2+2ab+\frac{bc}{m}=0}$$

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OR

Find the vertices and foci of the ellipse $\frac{(x+2{)}^2}{16}+\frac{(y-5{)}^2}{9}=1$.

Step 1: Rewrite the Ellipse in Standard Form

The given equation is: $$\frac{(x+2{)}^2}{16}+\frac{(y-5{)}^2}{9}=1$$ This is of the form: $$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$$ where:

• $(h,k)=(-2,5)$ is the center of the ellipse.
• $a^2=16⟹a=4$ (semi-major axis).
• $b^2=9⟹b=3$ (semi-minor axis).

Since $a>b$, the major axis is horizontal.

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Step 2: Find the Vertices

For a horizontal major axis, the vertices are at: $$(h\pm a,k)$$ Substitute the values: $$(-2\pm 4,5)$$ Thus, the vertices are: $$(-2+4,5)=(2,5)\text{and}(-2-4,5)=(-6,5)$$

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Step 3: Find the Foci

The distance of each focus from the center is given by: $$c=\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}$$ Thus, the foci are at: $$(h\pm c,k)=(-2\pm \sqrt{7},5)$$ So, the foci are: $$(-2+\sqrt{7},5)\text{and}(-2-\sqrt{7},5)$$

Final Answer:

• Vertices: $(2,5)$ and $(-6,5)$
• Foci: $(-2+\sqrt{7},5)$ and $(-2-\sqrt{7},5)$

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b) Show that the plane $2x+3y-4z=3$ is parallel to the plane $10x+15y-20z=12$ and is perpendicular to $3x+2y+3z=5$.

Step 1: Check Parallelism Between Two Planes

Two planes are parallel if their normal vectors are scalar multiples of each other.

• Plane 1: $2x+3y-4z=3$ has normal vector $\overrightarrow{n_1}=(2,3,-4)$.
• Plane 2: $10x+15y-20z=12$ has normal vector $\overrightarrow{n_2}=(10,15,-20)$.

Check if $\overrightarrow{n_2}=k\overrightarrow{n_1}$: $$\frac{10}{2}=5,\frac{15}{3}=5,\frac{-20}{-4}=5$$ Since all ratios are equal ($k=5$), the planes are parallel.

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Step 2: Check Perpendicularity with the Third Plane

A plane is perpendicular to another if its normal vector is perpendicular to the other plane's normal vector. This is checked using the dot product.

• Plane 1: $\overrightarrow{n_1}=(2,3,-4)$
• Plane 3: $3x+2y+3z=5$ has normal vector $\overrightarrow{n_3}=(3,2,3)$.

Compute the dot product: $$\overrightarrow{n_1}\cdot \overrightarrow{n_3}=(2)(3)+(3)(2)+(-4)(3)=6+6-12=0$$ Since the dot product is zero, the planes are perpendicular.

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Conclusion:

• The planes $2x+3y-4z=3$ and $10x+15y-20z=12$ are parallel because their normal vectors are scalar multiples.
• The plane $2x+3y-4z=3$ is perpendicular to the plane $3x+2y+3z=5$ because their normal vectors are orthogonal (dot product is zero).