Question

The line $x-y+4=0$ touches the ellipse $x^{2}+3y^{2}=12$ at ________.

• A. $(1, 3)$
• B. $(3, 1)$
• C. $(0, 2)$
• D. $(0, -2)$
• E. $(-3, 1)$

Hint:

Verify by checking if the point satisfies both the line and the ellipse equations.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
We need to find the point of tangency between a given line and an ellipse. We can solve this by substituting the line equation into the ellipse equation to find the single point of intersection, or by using the standard formula for the point of tangency.
Step 2: Key Formula or Approach
For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the point of tangency for a tangent line \(y = mx + c\) is given by the coordinates:
\[ \left( -\frac{a^2m}{c}, \frac{b^2}{c} \right) \] We first need to convert the given line and ellipse equations into these standard forms.
Step 3: Detailed Explanation
1. Convert the ellipse equation to standard form.
The ellipse equation is \(x^2 + 3y^2 = 12\).
Divide by 12 to make the right side equal to 1:
\[ \frac{x^2}{12} + \frac{3y^2}{12} = 1 \] \[ \frac{x^2}{12} + \frac{y^2}{4} = 1 \] From this, we identify \(a^2 = 12\) and \(b^2 = 4\).
2. Convert the line equation to slope-intercept form.
The line equation is \(x - y + 4 = 0\).
Rearranging to the form \(y = mx + c\):
\[ y = x + 4 \] From this, we identify the slope \(m = 1\) and the y-intercept \(c = 4\).
3. Verify the condition of tangency (optional but good practice).
The condition for the line \(y = mx + c\) to be tangent to the ellipse is \(c^2 = a^2m^2 + b^2\).
Left side: \(c^2 = 4^2 = 16\).
Right side: \(a^2m^2 + b^2 = (12)(1)^2 + 4 = 12 + 4 = 16\).
Since \(16 = 16\), the line is indeed tangent to the ellipse.
4. Calculate the point of tangency.
Using the formula \(\left( -\frac{a^2m}{c}, \frac{b^2}{c} \right)\):
x-coordinate: \(-\frac{a^2m}{c} = -\frac{(12)(1)}{4} = -\frac{12}{4} = -3\).
y-coordinate: \(\frac{b^2}{c} = \frac{4}{4} = 1\).
The point of tangency is (-3, 1).
Alternative Method (Substitution):
From the line equation, \(y = x+4\). Substitute this into the ellipse equation:
\(x^2 + 3(x+4)^2 = 12\)
\(x^2 + 3(x^2 + 8x + 16) = 12\)
\(x^2 + 3x^2 + 24x + 48 = 12\)
\(4x^2 + 24x + 36 = 0\)
Divide by 4: \(x^2 + 6x + 9 = 0\)
This is a perfect square: \((x+3)^2 = 0\).
This gives a single solution \(x = -3\), confirming tangency.
Now find y: \(y = x + 4 = -3 + 4 = 1\).
The point of tangency is (-3, 1).
Step 4: Final Answer
The point of tangency is (-3, 1).