Use the Lagrange multiplier method to find the shortest distance from the origin to
the hyperbola .

Ask by Carlson Adkins. in South Africa

Mar 23,2025

Question

Use the Lagrange multiplier method to find the shortest distance from the origin to the hyperbola $$ x^{2}+8 x y+7 y^{2}=225, z=0 $$.

UpStudy AI · Accepted Answer

We wish to minimize the distance from the origin to a point $$(x,y)$$ on the hyperbola 
$$
x^2+8xy+7y^2=225.
$$
Since the distance is minimized when its square is minimized, we set up the function to be minimized as
$$
D(x,y)=x^2+y^2.
$$

We introduce a Lagrange multiplier $$\lambda$$ and define the constraint function
$$
g(x,y)=x^2+8xy+7y^2-225=0.
$$
According to the method of Lagrange multipliers, we must have
$$
\nabla D = \lambda\, \nabla g.
$$

**Step 1. Compute Gradients**

First, compute the gradient of $$D(x,y)$$:
$$
\frac{\partial D}{\partial x}=2x,\quad \frac{\partial D}{\partial y}=2y.
$$
Thus,
$$
\nabla D = (2x,\, 2y).
$$

Next, compute the gradient of $$g(x,y)$$:
$$
\frac{\partial g}{\partial x}=2x+8y,\quad \frac{\partial g}{\partial y}=8x+14y.
$$
So,
$$
\nabla g = (2x+8y,\, 8x+14y).
$$

**Step 2. Set Up the Lagrange Equations**

We require:
$$
\begin{cases}
2x = \lambda (2x+8y), \
2y = \lambda (8x+14y), \
x^2+8xy+7y^2=225.
\end{cases}
$$

Divide the first two equations by 2 to simplify:
$$
\begin{cases}
x = \lambda (x+4y), \quad \text{(1)}\[1mm]
y = \lambda (4x+7y). \quad \text{(2)}
\end{cases}
$$

**Step 3. Expressing Relations**

Rearrange equation (1):
$$
x = \lambda x + 4\lambda y \quad \Longrightarrow \quad x - \lambda x = 4\lambda y,
$$
which gives
$$
x(1-\lambda)=4\lambda y.
$$
Assuming $$x$$ and $$y$$ are not zero (we will consider the zero cases later), we solve for $$y$$:
$$
y=\frac{1-\lambda}{4\lambda}\,x. \quad \text{(3)}
$$

Now rearrange equation (2):
$$
y = 4\lambda x + 7\lambda y \quad \Longrightarrow \quad y - 7\lambda y = 4\lambda x,
$$
or
$$
y(1-7\lambda)=4\lambda x.
$$
Solving for $$x$$ (again assuming $$y\neq 0$$):
$$
x=\frac{1-7\lambda}{4\lambda}\,y. \quad \text{(4)}
$$

**Step 4. Find $$\lambda$$ by Consistency**

Substitute the expression for $$y$$ from (3) into (4):
$$
x = \frac{1-7\lambda}{4\lambda}\left(\frac{1-\lambda}{4\lambda}x\right).
$$
Assuming $$x\neq 0$$ we cancel $$x$$ to obtain:
$$
1=\frac{(1-7\lambda)(1-\lambda)}{16\lambda^2}.
$$
Multiply both sides by $$16\lambda^2$$:
$$
16\lambda^2 = (1-7\lambda)(1-\lambda).
$$
Expand the right-hand side:
$$
(1-7\lambda)(1-\lambda) = 1 -\lambda -7\lambda +7\lambda^2 = 1-8\lambda+7\lambda^2.
$$
Thus, the equation becomes:
$$
16\lambda^2 = 1-8\lambda+7\lambda^2.
$$
Subtract $$1-8\lambda+7\lambda^2$$ from both sides:
$$
16\lambda^2-7\lambda^2+8\lambda-1=0,
$$
which simplifies to:
$$
9\lambda^2+8\lambda-1=0.
$$

Solve the quadratic equation using the quadratic formula:
$$
\lambda=\frac{-8\pm\sqrt{8^2-4\cdot9\cdot(-1)}}{2\cdot9}=\frac{-8\pm\sqrt{64+36}}{18}=\frac{-8\pm\sqrt{100}}{18}.
$$
Thus,
$$
\lambda=\frac{-8\pm10}{18}.
$$

This gives two solutions:
$$
\lambda_1=\frac{-8+10}{18}=\frac{2}{18}=\frac{1}{9},\qquad\lambda_2=\frac{-8-10}{18}=-1.
$$

**Step 5. Investigate Each $$\lambda$$ Value**

*Case 1: $$\lambda=\frac{1}{9}$$*

Using equation (3):
$$
y=\frac{1-\frac{1}{9}}{4\cdot\frac{1}{9}}\,x=\frac{\frac{8}{9}}{\frac{4}{9}}\,x=\frac{8}{4}\,x=2x.
$$
So we have:
$$
y=2x.
$$

Substitute $$y=2x$$ into the constraint:
$$
x^2+8x(2x)+7(2x)^2=225.
$$
Simplify the above:
$$
x^2+16x^2+28x^2=225\quad\Longrightarrow\quad45x^2=225.
$$
Solve for $$x^2$$:
$$
x^2=\frac{225}{45}=5.
$$
Thus,
$$
x=\sqrt{5} \quad \text{or} \quad x=-\sqrt{5},
$$
and correspondingly,
$$
y=2\sqrt{5} \quad \text{or} \quad y=-2\sqrt{5}.
$$

The squared distance from the origin is:
$$
D(x,y)=x^2+y^2=5+4\cdot5=5+20=25,
$$
so the shortest distance is:
$$
\sqrt{25}=5.
$$

*Case 2: $$\lambda=-1$$*

Using equation (3) again:
$$
y=\frac{1-(-1)}{4(-1)}\,x=\frac{2}{-4}\,x=-\frac{1}{2}x.
$$
Substitute $$y=-\frac{1}{2}x$$ into the constraint:
$$
x^{2}+8x\left(-\frac{1}{2}x\right)+7\left(-\frac{1}{2}x\right)^{2}=225.
$$
Simplify term by term:
$$
x^{2}-4x^{2}+7\frac{x^{2}}{4}=x^{2}-4x^{2}+\frac{7x^{2}}{4}.
$$
Combine the terms:
$$
x^{2}-4x^{2}=-3x^{2},
$$
so we have:
$$
-3x^{2}+\frac{7x^{2}}{4}=\frac{-12x^{2}+7x^{2}}{4}=\frac{-5x^{2}}{4}=225.
$$
This implies:
$$
x^{2}=-\frac{225\cdot4}{5}=-180,
$$
which is not possible for a real $$x$$. Hence, $$\lambda=-1$$ does not yield a valid solution.

**Step 6. Conclusion**

The only valid solution corresponds to $$\lambda=\frac{1}{9}$$, giving the points:
$$
(x,y)=(\sqrt{5},2\sqrt{5})\quad \text{or} \quad (x,y)=(-\sqrt{5},-2\sqrt{5}).
$$
The shortest distance from the origin to the hyperbola is:
$$
\sqrt{x^2+y^2}=\sqrt{25}=5.
$$
The shortest distance from the origin to the hyperbola $$ x^{2}+8xy+7y^{2}=225 $$ is 5 units.

Use the Lagrange multiplier method to find the shortest distance from the origin to
the hyperbola .

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Use the Lagrange multiplier method to find the shortest distance from the origin to the hyperbola .