Question Statement

Find the value of $m$ ($m\ne 0$) such that the line $5x+my-20=0$ touches the ellipse $\frac{x^2}{4}+\frac{y^2}{5}=1$. Also, find the point where it touches the ellipse.

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Background and Explanation

To find the condition for a line to be tangent to a curve, we substitute the equation of the line into the equation of the curve to form a quadratic equation. A line is tangent to a curve if they intersect at exactly one point, which mathematically means the discriminant ($D=b^2-4ac$) of the resulting quadratic equation must be equal to zero.

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Solution

Given the equation of the line: $5x+my-20=0⟹my=20-5x⟹y=\frac{5(4-x)}{m}$

And the equation of the ellipse: $\frac{x^2}{4}+\frac{y^2}{5}=1$

Step 1: Substitute the line into the ellipse equation

Substitute the value of $y$ from (i) into (ii): $\frac{x^2}{4}+\frac{{\left(\frac{5(4-x)}{m}\right)}^2}{5}=1$ $\frac{x^2}{4}+\frac{25(4-x{)}^2}{5m^2}=1$ $\frac{x^2}{4}+\frac{5(4-x{)}^2}{m^2}=1$

To clear the denominators, multiply the entire equation by $4m^2$: $m^2{x}^2+20(4-x{)}^2=4m^2$ $m^2{x}^2+20(16-8x+x^2)=4m^2$ $m^2{x}^2+320-160x+20x^2=4m^2$

Rearrange into the standard quadratic form $a{x}^2+bx+c=0$: $(m^2+20)x^2-160x+(320-4m^2)=0$

Step 2: Apply the condition for tangency

The line will touch the ellipse if the discriminant of equation (iii) is zero ($D=0$): $(-160{)}^2-4(m^2+20)(320-4m^2)=0$ $25600-16(m^2+20)(80-m^2)=0$

Divide both sides by 16: $1600-(m^2+20)(80-m^2)=0$ $1600-(80m^2-m^4+1600-20m^2)=0$ $1600-(60m^2-m^4+1600)=0$ $1600-60m^2+m^4-1600=0$ $m^4-60m^2=0$ $m^2(m^2-60)=0$

Since the problem states $m\ne 0$: $m^2-60=0⟹m^2=60$ $m=\pm \sqrt{60}=\pm 2\sqrt{15}$

Step 3: Find the points of contact

Case 1: When $m=-2\sqrt{15}$ Substitute $m^2=60$ into equation (iii): $[60+20]x^2-160x+(320-4(60))=0$ $80x^2-160x+(320-240)=0$ $80x^2-160x+80=0$ Divide by 80: $x^2-2x+1=0⟹(x-1{)}^2=0⟹x=1$

Now find $y$ using $y=\frac{5(4-x)}{m}$: $y=\frac{5(4-1)}{-2\sqrt{15}}=\frac{15}{-2\sqrt{15}}=-\frac{15\sqrt{15}}{2(15)}=-\frac{\sqrt{15}}{2}$ The first point of contact is $\left(1,-\frac{\sqrt{15}}{2}\right)$.

Case 2: When $m=2\sqrt{15}$ Substitute $m^2=60$ into equation (iii): $80x^2-160x+80=0⟹x=1$

Now find $y$: $y=\frac{5(4-1)}{2\sqrt{15}}=\frac{15}{2\sqrt{15}}=\frac{\sqrt{15}}{2}$ The second point of contact is $\left(1,\frac{\sqrt{15}}{2}\right)$.

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Key Formulas or Methods Used

• Substitution Method: Solving systems of equations by substituting one variable into another equation.
• Condition for Tangency: For a quadratic equation $a{x}^2+bx+c=0$, the line is tangent to the curve if the discriminant $D=b^2-4ac=0$.
• Standard Form of Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

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Summary of Steps

1. Isolate $y$ in the line equation in terms of $x$ and $m$.
2. Substitute this $y$ expression into the ellipse equation.
3. Simplify the resulting expression into a standard quadratic equation in $x$.
4. Set the discriminant of the quadratic equation to zero to find the values of $m$.
5. Solve for $m$ and substitute these values back into the quadratic equation to find the $x$-coordinate of the point of contact.
6. Substitute $x$ and $m$ back into the linear equation to find the corresponding $y$-coordinates.